Integral Calculus for Beginners
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INTEGRAL
CALCULUS
FOR
BEGINNERS
CALCULUS
INTEGRAL FOR
WITH
AN
BEGINNERS
INTRODUCTION
THE
TO
DIFFERENTIAL
STUDY
EQUATIONS
BY
JOSEPH FORMERLY
FELLOW
M.A.
EDWAEDS, OF
SIDNEY
SUSSEX
MACMILLAN AND
COLLEGE,
AND NEW
YORK
1896 All
rightsreserved
CAMBRIDGE
CO.
OF
First
Reprinted With
and
additions
GLASGOW
:
BY
PRINTED
ROBERT
Edition, 1891,
1890.
corrections,
AT
MACLEHOSE
1893.
1892, 1894
THE
;
reprinted
UNIVERSITY
AND
1896.
PRESS
CO.
PREFACE.
THE
introduction
to
for
suitable
a
a
it does
therefore
the
of
omission
It
will
be
of
all
found,
and
calculation
of
solids
of
afforded the
to
Centroid, As in want
of
it
or
seems
Applied of
the
in
and
Some of
such
of
undesirable
the
path
with
the
M298720
be methods
also
method
position of
should
is
applications
Moment
that
Mathematics
acquaintance
a
the
surfaces
general
the
obtaining
value
also
as
and
the
cesses pro-
Quadrature,
useful
as
which
reading.
indication
other
rather
ordinary
volumes
the
student
employed
later
a
the
Rectification
but
subject
fully treated,
are
Integral Calculus,
be
to
for
that
its
for Beginners,
the
left
however,
revolution.
the
of
Like
subject.
completeness,
at
best
of
methods
the
the
portions
as
integration
principal
of
aim
sound
a
Integral Calculus,
the
beginning
usually regarded
are
of
form
to
Calculus 'Differential!
the
at
intended
study
student
companion, not
is
volume
present
of
a
Inertia. of
a
blocked of
student
by
a
solving
vi
PREFACE.
elementaryDifferential Equations,and time
that
his
of
study and
exhaustive
added
of
the
includingall with
meet
Linear
such
kinds
his
reading
consistent The been
subjecthas with
the scope
carefullymade of
the student the
"
treated
attackingthe
of the
and originality
A
able consider-
be
explained
harder
ingenuity,though
in
to
been
actuallyset
which
usuallyindicated.
I
am
in
the
sets
at
generallyof
are
character, and
worked
call for few
a
greater
present any
A largeproportionof difficulty.
examples have sources
somewhat
have
text
firmly fixed
chapters.These
miscellaneous
considerable
be
is
illustrate the
to
examples should
may
cients, Coeffi-
present work.
selected
"
general
fullyas
as
that the several methods
so
book-work
before ends
these
to and
Constant
with
of the
or
of the
elementaryparts of
they immediatelyfollow.
number
mind
been
been
likelyto
is
examples scattered throughoutthe
articles which
the
solution
solution of the
Equation
tematic sys-
Analytical Statics,
the
to the
a
has
account
student
of
same
complete
some
of
the
as
Particle,and
a
Differential the
more
brief
ordinarymethods
Rigid Dynamics. Up
the
subject in
the
treatise,a
in
of
Dynamics
in
stopped for
elementaryforms occurring, leadingup
more
by
be
should
course
at the
these
examinations, and
indebted
for them
are
PREFACE.
acknowledgments
My
of
works
the
of
Treatises
of, Bertrand
Greenhill's
which
Calculus, consult
with
My
thanks
kindly the
of
many
sent
great
me
desirable
1894.
the
especially
to
the
Chapter
the
advanced
more
to
on
Todhunter,
and
to
the
on
fessor Pro-
Integral
student
may
advantage.
and
to
friends
several
suggestions
plan
of
the
JOSEPH
October,
degree
some
writers
interesting
valuable
scope
more
and
due
are
modern
the
but
in
due
are
treated
subjects
vii
who
with
regard
work.
EDWARDS.
have
to
CONTENTS.
INTEGRAL
CALCULUS.
CHAPTER
NOTATION,
I.
SUMMATION,
APPLICATIONS. PAGES
Determination
of
1
Area,
an
3 "
......
Integration Volume
of
from
the
"
13
10
Revolution,
"
CHAPTER
GENERAL
Fundamental
9
4
Definition,
II.
FORMS.
STANDARD
METHOD.
19
14
Theorem,
"
.......
Nomenclature
and
21
20
Notation,
"
......
General
obeyed
Laws
by
the
Integrating
22
Symbol, .
Integration Table
of
of
xn,
23"26
x~l,
26"28
Results,
CHAPTER
METHOD
Method The
of
Changing
Hyperbolic
.
the
III.
OF
Variable,
SUBSTITUTION.
29
32 "
33
Functions,
36 "
......
Additional
Standard
Results,
37"41
CONTENTS.
IV.
CHAPTER INTEGRATION
PARTS.
BY
PAGES
Geometrical
of
Parts"
Integration"by
Product,
a
....
48"49
Proof, of the
Extension
50"52
Rule,
V.
CHAPTER
FRACTIONS.
PARTIAL Standard
55
Cases,
57
"
........
Fraction
General
43"47
nominator, De-
and
Numerator
Rational
with
58"61
VI.
CHAPTER
Integrationof
METHODS.
STANDARD
SUNDRY
f^L
65"68 .
...
J \/K of Sines and
Products
Powers
and
Powers
of Secants
Powers
of
or
Tangents
+ o
cos
77
,
79"83
x
CHAPTER
VII.
REDUCTION
Integrationof xm-lXP, Reduction
Formulae
Reduction
Formulae
where
for
for
X
Evaluation
of
FORMULAE.
=
a
+
bxn,
/ xm~lXpdx,
/ sii\nxdx,
.
.
.
87"89 90
93
"
....
/ sin^a; cos^a: dx,
.
.
.
.
.
94
"
95
IT
7T
j"z
78
"
.....
etc.,
"
76
"
.....
Cotangents,
or
69"74
.
.
.
75
Cosecants,
/rfv a
Cosines,
r -i
I
ain^x cos^x
dx,
.
96
"
102
CONTENTS.
xi
CHAPTER
VIII.
MISCELLANEOUS
METHODS. PAGES
/^).fx.9 X. Y
Integrationof
109"117
.......
J
i\f
Integrationof some SpecialFractional Forms, General and Geometrical Illustrations, Propositions Some Elementary Definite Integrals, Differentiation under an IntegralSign, .
118
119
"
.
120
"
124
.
125
....
128
....
"
"
127 129
IX.
CHAPTER RECTIFICATION. Rules
for
Formulae
Curve-Tracing,
of
for
Equation,
of Pedal
Illustrative
Closed Curve,
a
Evolute,
an
Intrinsic Arc
and
for Rectification
Modification Arc
135"137
.......
Examples,
.
13S
"
140
.....
143
........
144
........
Curve,
139
"
149 150
........
CHAPTER
X.
QUADRATURE. Cartesian
Formula,
Sectorial Areas. Area
of
Other Area
a
Curve,
a
of
'.''-. .
two
of Curvature
Radii
the
168"175
........
176
.......
CHAPTER
of
164"165
166"167
CorrespondingAreas,
Volumes
and
.........
Pedals,
SURFACES
157
161"163
.......
.....
Curve,
"
158"160
.......
Expressions, Evolute,
Areas
Polars,
Closed
between
153
........
AND
VOLUMES
Revolution, Surfaces of Revolution,
"
177
XI. OF
SOLIDS
OF
REVOLUTION.
.......
183"184
.......
185"187
CONTENTS.
xii
PAGES
Theorems
of
Revolution
Pappus,
of
a
188
.......
Sectorial
192
Area,
......
XII.
CHAPTER ELEMENTS
SECOND-ORDER
191
AREA.
OF
MISCELLANEOUS
APPLICATIONS. Surface
Cartesian Integrals, of
; Moments
Centroids
Element,
Inertia,
195
.
199
.....
Polar Element, Integrals, Centroids,etc.,Polar Formulae,
204
.....
DIFFERENTIAL
of
a
Differential
207
"
EQUATIONS. XIII.
CHAPTER
Genesis
201
"
202"203
Surface
EQUATIONS
198
"
OF
Equation,
ORDER.
FIRST
THE
211"214
.....
215
Variables Linear
Separable, Equations,
216
XIV.
CHAPTER
EQUATIONS
OF
ORDER
FIRST
THE
(Continued}. 221
Homogeneous Equations, One
Letter
Clairaut's
226
"
227"229
Absent,
230"233
Form,
XV.
CHAPTER
EQUATIONS
219
"
OF
THE
EXACT
ORDER.
SECOND
DIFFERENTIAL
EQUATIONS. Linear One
Letter
General Exact
235
Equations,
"
236
237"238
Absent,
Equation. Removal Differential Equations,
of
Linear
.
.
Term,
a
.
.
239 .
.
.
.
"
240
241"242
CONTENTS.
xiii
CHAPTER
XVI.
EQUATION
DIFFERENTIAL
LINEAR
CONSTANT
WITH
COEFFICIENTS. PAGES
General
of
Form
The
Complementary
The
Particular
245
Function,
252"263 Constant
with
Form
Linear
to
264"265
Coefficients,
XVII.
CHAPTER ORTHOGONAL
Some
Important
266"269
Equations,
Dynamical
Illustrative
Further
EQUATIONS.
MISCELLANEOUS
TRAJECTORIES.
Trajectories,
Orthogonal
251
"
.....
Integral, Reducible
Equation
An
243"244
Solution,
270
271
"
....
272"277
Examples,
278"308
Answers,
ABBREVIATION. To
indicate
derived, in in
cases
the
common,
(a)
the
=
St.
where
which
from
sources
a
group
references
are
of
colleges
abbreviated
Peter's, Pembroke,
Corpus
of
many
examples
held
an
follows
:
have as
the
are
examination "
Christi, Queen's,
and
St.
Catharine's.
(j8) =
Clare, Caius, Trinity Hall, and
(7)
=
Jesus, Christ's, Magdalen,
(d)
(e)
=
=
Jesus,
Christ's, Emanuel,
Clare, Caius, and
King's.
King's.
Emanuel, and
Sidney
and
Sidney
Sussex.
Sussex.
CALCULUS
INTEGRAL
CHAPTER
NOTATION,
The
of
area
of
Aim
obtain
the
the
Calculus
Integral to
APPLICATIONS.
SUMMATION,
and
Use
1.
I.
the
is
plane
method
of
bounded
space
of
outcome
general
some
Calculus.
Integral
by
an
deavour en-
the
finding
curved
given
lines. In
the
area
it is
into
a
then
the
limit is
each
number
of
is
problems
such
the
volumes
bounded
moments
of
E.
i.
c.
that
the of
areas
up
obtaining
elements small
when
and
their
increased. when
discovered, as
of
these
an
elements.
small
very
infinitesimally
found
be
all
such
divided
space
method
some
of
sum
ultimately
summation
line,
the
of
form
to
infinitely will
It
have
this
suppose
number
large
of
determination
the
to
necessary
very
We
of
problem
inertia,
it
finding surfaces
by the
be
may
of of
them,
the
length
the
of
method
a
applied
given
positions A
such
once
of
shape
other
to a
curved
and
the of
determination of
Centroids,
etc. "E
CALCULUS.
INTEGRAL
2
Throughout the book all coordinate all angles will supposed rectangular, measured
in
circular
measure,
supposed Napierian, exceptwhen of 2. Determination Notation. be Summed.
an
Area.
and
axes
will be
be supposed all logarithms stated.
otherwise
of Series to
Form
of the portion to find the area Supposeit is required of space bounded AB, defined by by a given curve of and BM its Cartesian equation, the ordinates AL A
and
B, and the cc-axis.
L
0,0,0,0, Fig. 1.
Let
LM
QiQz,Q^Qv
be divided into n equal small "f lengthA, and let eacn """"
parts,LQV
and 6 be ?i/L Also if the ordinates a
Then b abscissae of A and ". a be the equationof the curve, (f)(x) y LA, QiPp $2^2*e^c-'through the several the =
"
=
points L, ^(a+K), ^(a+2A),etc. Qv Q2,etc.,are of lengths"j"(a), Let their extremities be respectively A, P1? P2, etc., and completethe rectangles AQV PjQg,P2Q3,etc. of these n rectangles falls short of Now the sum of the n small figures, the area sought by the sum etc. Let each of these be supposed 1, P1J22P2,
CALCULUS.
INTEGRAL
the
term
h(f"(a+nh) or
the
limit
is taken.
which
A0(6)
Hence
vanishes of
limit
the
when
this
series
also be written
may
f6 I
"t"(x)dx.
a
3.
This
summation as
be
sometimes
may
elementary means,
Definition.
the
Integration from
to illustrate
proceed
now
we
by
effected : "
Cb
Ex.
Here
1.
we
/ e*dx.
Calculate
have
evaluate
to
+ ea+"
+ ea+h Lth==Qh[ea
b
where
=
a
+
.
.
.
+ nh.
ea)-^-=e* =Lth^h^p\ea=Lth^(eb
This
-
-
1
"
"
e\
X
"
[By Diff. Calc. for Beginners,
Art.
15.]
r=n-l
/b
xdx
we
to find
have
2
Lt
("+rA)A,
where
r="
2(a + rh)h
Now
and
in the
=
limit
becomes
22'
2
/61 "$x a
diminished indefinitely
of
we
have
to
obtain
the
limit
when
h is
APPLICATIONS.
SUMMATION,
NOTATION,
5
"
b+ h
a
"'
a-h and
when
h diminishes
without
limit,each
of these becomes
II
b'
a
/"==*.* f*JL
Thus
J
.r2
b
a
a
Ex. 4.
Prove
ab initio that
/"
sin
We
now
ofo?
#
cos
=
a
6.
cos
-
to find the limit of
are
[sina
+
sin(a+ k)+ sin(a+ 2A)+
sinf a+n
"
\
sin
.
.
.
to
terms]A,
n
l- Jsin n2/ 2,
| *
This
expression =
cosfa
" -
J
cos
"
" a
+
(2n
-
1)-j"
-
2JJsin2
sm-
which
when
A is
small ultimately takes indefinitely cos
a
"
cos
b.
the form
CALCULUS.
INTEGRAL
EXAMPLES. Prove
by
that
summation
/ sinh sir xdx /
2.
cosh b
"
cosh
"
a.
3.
/b
OdO
cos
4.
Integrationof
As
a
further
sin 6
=
a.
xm.
example we
the limit of the
sin
"
next
propose
to consider
+ 1 not
being zero.
of the series
sum
h[am+ (a+ h}m+ (a+ 2h)m+. 6
i
h7
where
=
a
"
--
,
n
and
n
is made
[Lemma.
The
"
m indefinitely large,
Limit
of
fy v"/
I\m+l
I
yin +
_
%
2
-
"
1
is
m
+ 1
when
A
is
Aym
diminished,whatever indefinitely y finite magnitude. For the expression be written may
may
be, provided it
be
of
-1
y and
since h is to be
ultimatelyzero
we
may
consider
-
to
be
y
less
than
unity,and
we
Theorem
to
expand
therefore
may
/
^\7?l+ ( 1 -J--
J
apply
the
Binomial
l ,
whatever
be the value
of m+l.
(See Dif.
Gale,
APPLICATIONS.
SUMMATION,
NOTATION,
for Beginners,Art.
Thus
13.)
the
7
expression
"becomes
-x(a convergent series) y A is
+ I when
"m
diminished.] indefinitely
In the result
put
i/ success! vely
and
we
a+h, a+2h,etc....a + (n
a,
=
"
l)h,
get
l-am+l_
(
T
~
r, _
1 -
(a + n^
_
h(a+n-Ui)m for
adding numerators
or
for
a
numerator
new
a
and
nominat de-
denominator,
new
+ (a+ /t)w + (a + 2h)m+ fe[aw
.
.
.
+
(a+ n^l
or
+ (a+ A)m+ (a Lth=Qh[am
'
m+1 In may
accordance
with
the
notation
be written '6
b" 7
xmdx="
m+1
of Art.
2, this
INTEGRAL
8 The
letters
CALCULUS.
and b may providedxm does not
When
whatever, represent any finite quantities
a
is taken is necessary in the a
become
small exceedingly to proof suppose h an
6
=
1 and
xmdx=
a
"
in the
that
limit
if
"7
-
is
zero
for
V
0, ultimatelythe
=
and
x=a
and ultimately zero, it infinitesimal of higher
as
order,for it has been assumed all the values givento y. When
infinite between
+ 1 be
m
comes be-
theorem
positive,
o
or
This theorem
if m
oo
=
+ 1 be
be written
may
negative.
also
"r
as according
or,
which
is the
oo
the
"M4-'
former
by 1
and
limit
thing,
same
-Lstn=
differs from
negative.The
is positive or
m+1
is therefore also
"
-"?,
positiveor negative. The
i.e. by 0 in the
"
,
limit,
n or
oo
case
as m+1 according
when
m
+ l=0
is will
be discussed later. Ex.
bounded
1.
Find
by
the
area
the curve,
of the the
the parabola 7/2=4a# the ordinate x"c.
portionof
#-axis,and
NOTATION, Let
length c into n equal portionsof which Then if (r+l)th,and erect ordinates NP, MQ.
divide
us
is the
NM
PR when
the
is infinite of the
n
sum
Lt^PN.NM
i.e. where
nh
the
parallelto NM,
drawn
be
APPLICATIONS.
SUMMATION,
of such
required is the limit (Art. 2), rectanglesas PM area
or
c.
=
Now
[By Area
of the
=f
of the Ex.
Find
2.
the
4.]
=f
rectangleof which the are adjacentsides.
area
Art.
mass
of
a
extreme
rod
whose
ordinate
arid abscissa
densityvaries
the
as
of the distance from one end. with power Let a be the length of the rod, o" its sectional area supposed uniform. Divide the rod into n elementary portions each of
length of
The
-.
densitv
zero
is w-,
\
"
-
"
n
and
(r+l)th
its
element
from
density varies from
n
(7+la\m 1
of the
volume
)
Its .
mass
is therefore
coa**1-
( *
intermediate
)
and
**
between
end
the |
"
*"
to
INTEGRAL
10 Thus
the
mass
of the whole
CALCULUS.
rod lies between
and and
in the
limit,when
n
increases
becomes indefinitely,
ra+1
5. Determination
of
a
Volume
of Revolution.
requiredto find the volume formed by the revolution of a given curve about an axis AB in its own planewhich it does not cut. Taking the axis of revolution as the cc-axis,the figuremay be described exactlyas in Art. 2. The Let
it be
Fig. 3.
elementaryrectangles AQV P-fy^P2Qz"etc.,trace in and their revolution circular discs of equal thickness, of volumes "jrAL2 LQ19nrP^ Q", etc. The several annular portionsformed by the revolution of the portionsAR^^ P^R^P^ P2E3P3, etc.,may be con.
.
CALCULUS.
INTEGRAL
12
before
dividingas
Then
into
elementarycircular laminae, we
have
/cy^dx
re
/ xdx
4a:r
"
2
=J cylinderof [Or if expressedas Volume
I
4a?r
AN
radius
PA7' and
heightAN.
series
a
[c
=
[Art.4.]
dx
x
o
r = "
.
2a?rc2.]
[Art.4.]
2
Ex.
2.
revolution
Find
the volume
of the
prolatespheroidformed
of the
ellipse~+^2
=
by
the
the #-axis.
l about
*
Fig. 5.
Dividing as
before
coincide with
axes
the
into
elementary #-axis,the volume
circular laminae is twice
/ Try^dx. Now
-a2
-
x*)dx
a
which, accordingto Article 4, is equal to
5[a*.("-0)-^] or
and
the whole
volume
is
whose
APPLICATIONS.
SUMMATION,
NOTATION,
[or if desirable we may the sign of integration, as
obtain
the
without
result
same
13
using
EXAMPLES. 1. Find
the
the
ordinates
2. If the
volume
of the
formed
volume
the the
volume
x2+y2 the
of the
a2 about
=
of
areas
point
revolution the
2, the
the
the
by
the
as
the
the
the
of the
the
line
x
this
when
triangle =
a.
triangle
revolution
of
.r-axis.
figures bounded the
area
each
by
ordinate
of each
x
=
about
of
the
h ; also the the #-axis :
=
aty a
circular
distance the
disc from
of
point to
the
density at
centre.
prolate spheroid =
each
which
the
l ellipse^2/a2-f^/2/62
density at
the
by
7/3 a*a
of
mass
the
revolution
of
mass
varies
of
area
sphere formed
the
(8) 7. Find
,#-axis find
the
by the reel-shapedsolid formed that part of the parabolay^"^ax
#-axis, and
the
curves,
formed
8. Find
of Art.
y-axisof
the
(a)
each
round
1 revolve
of the
volume
latus-rectum.
6. Find
following
^^ the #-axis,and
"
#-axis.
the
about
5. Find
circle
method
the
the
revolution
the
y
solid formed.
about
by
curve
line y=x tan 0, the #-axis and of the cone formed also the volume
4. Find
off
the
the
Find
cut
Question
in
by
by
revolves
by
#=".
#=a, area
3. Find
bounded
area
about
be //x,
formed the
by the #-axis,supposing
II
CHAPTEE
METHOD.
GENEEAL
Before
6. the
theorem the
Calculus,
of
the
with
indicated
operation
of
general
a
enable
cases
many
applications
establish
shall
we
in
will
which
result
further
proceeding
Integral
FOEMS.
STANDAED
to
us
infer
by
n
I
"p(x)dx
a
without
recourse having difficult,process
often
the
to
of
tedious,
usually
Algebraic
and
Trigonometrical
or
Summation. 7. finite
b
difference
h,
so
limit
of
ike
h
when increased
[It be
the
may
that
b
a
"
of
sum
be
once
greatest
term
6,
"
into It
is
and
n
suppose
portions
required
find
to
a
the each
the
4"(a + 2h)+...
+
indefinitely,
0(6 and
-
h)
+
0(6)],
therefore
n
limit.
without at
be
a
values
series
diminished
is
let
nh.
=
the +
;
is
which
x
finite
given
divided
be
"p(a + h)
+
x
to
a
"
equal
ft[0O)
variable
the
of
function
any
between
continuous
and b of
and
"/)(x) be
Let
PROP.
seen
the
-
that sum
this
limit
is
is
a)"t"(a+ rh)
+
h"$"(a+
finite, for
if
"$"(a+rh)
x
b and
between
intermediate
Let
15
be \fs(x)
of
function
another
shall then
of
a.]
i.e.such its differential coefficient,
We
FORMS.
is finite four all values since by hypothesis""(#) finite,
is
which
STANDARD
METHOD.
GENERAL
x
such
is "j)(x)
that
that
that
prove
Lth^["fa)+^a+h)+^ By
definition
and
therefore
where
a:
a
=
limit
quantity whose indefinitely ; thus
is
diminishes
^a)*
a
is
zero
+halt
=\/s(a+ 7i) t/r(a)
h(j)(a) Similarly h"f"(a
h
when
"
etc., + nil) Ih) \[s(a =
"
where
the
"
quantitiesa2,
quantitieswhose
limits
a3,
...,
zero
are
\
an
all, like
are
av
h diminishes
when
indefinitely. By addition, + 0(a + h)+ "f"(a h["f"(a)
Let
a
be the
greatestof
the
quantitiesav
a2,
.
.
.
then
Afoi+ag+^.+On] and
therefore
vanishes
is
"nha,
i.e.
in the limit.
"("
Thus
"
a)a,
,
an,
CALCULUS.
INTEGRAL
16
limit zero; hence if desire,it may be added to the left-hand member this result,and it may then be stated that The
term
is in fc"/"(6)
the
we
of
1 "/)(x)dx\ls(b)\ls(a).
.e.
=
result
This
"
denoted is frequently \[s(b)\fs(a)
the
the form
of
p\/r(a3) J
notation
.
this result it appears function ^fs(x) (of which
that when
From the
by
"
is obtained, the coefficient)
is "/"(x)
of
process
differential
the
algebraicor
rb
summation trigonometric
I
obtain
to
avoided. The
"j)(x)dx may
be
a
supposed in finitequantities.We shall
the above work extend our now
I "f"(x)dx express
the limit when
letters b and
to denote
notation
to let
as
so
a
are
a
b becomes
i.e. infinitely largeof ^(6) ^js(a), "
I
(j)(x)dxLtb=x I "j)(x)dx. =
a
a
(j)(x)dx fb
understood
shall be
we
to
fb
-\I,(a)]
or
Ex. Hence
1.
if
The
and "df(x)":L^ ' m+l
coefficient of ^"
differential
we "$"(x)=xm
Lta=00\"f"(x)dx.
-
is
plainlyxm.
have
/
J
x
m+1
m
+ \
m+l
STANDARD
METHOD.
GENERAL
Ex. 2. The quantitywhose Hence known to be sin x.
FORMS.
17
differential coefficient is
cos
a?
is
"6
sin b
=
quantitywhose
The
Ex. 3. itself ex.
dx
x
cos
sin
"
a.
differential coefficient
is e* is
Hence
Ex. 4.
EXAMPLES. the values of
down
Write
rl
/V"dr, CiX) /b
1.
,-2
/V"cfo?, 3.
2. I X 2i,
X
Cf/JCm
0
a
I
o.
cfo, d/X^
X
4.
1
it
ir
/2 cos
rA
dx,
x
6.
/
r4
sec2^;dx^
7.
/
\
o
ia
8. Geometrical The
proof of thus be a
the
Illustration of Proof. above
theorem
may
be
interpretedgeometricall
:
"
of which the ordinate is finite Let AB portionof a curve and continuous all points between A and B, as also the at makes tangent of the angle which the tangent to the curve with the a?-axis. Let the abscissae of A and B be a and b respectively. Draw ordinates A N, BM. be divided Let the portionNM into n equal portionseach of lengthh. Erect ordinates at each of these pointsof division
cutting the
curve
tangents AP^
in
P, Q, R,
PQi, QRi, etc.,and
Draw lines
etc.
...,
the
the
successive
AP2,PQ2JQR2,..., parallelto
^r(x\and
y then =
the
and ,r-axis,
let the
equation of
the
V^') ""M" + Zh\ etc., are + h\ "$"(a "f"(a\ "$"(a respectively let
=
tanP.JPj, E.
I. C.
taii^Pft, B
etc.,
curve
be
CALCULUS.
INTEGRAL
18 and
-h),
...,
it is clear that the
Now
P2P, "2", R2R,
the lengths respectively
are
of
sum algebraic
is ...,
i.e.
MB-NA,
Hence
s,
u
L
M
x
Fig. 6.
portion within square brackets may be shewn for instance be For if R^ with h. diminish indefinitely PjP, Q^ etc.,the sum greatestof the several quantities the
Now
[P1P+Q1Q+...] if the abscissa of
But
Q
is
"nR1R,
i.e.
to
the
"(b-a)-}~.
be called #, then
(x)+
and
-^"(x+
Qh\
[Diff.Calc. for Beginners, Art. 185.] so
R^R
that
"(x4- "9A) =
(x+ Oh),
(6 a)
and which
=
-
is
an
infinitesimal in
generalof
the first order.
20
CALCULUS.
INTEGRAL
10. When we
are
lower
the
limit
is not specified and merely enquiringthe form of the (at present) a
function \fr(x), unknown differential coefficient whose is the known function $(#), the notation used is
the limits
beingomitted.
11. Nomenclature. The
of these
nomenclature
is as expressions
follows
:
b
r(p(x)dx is called the "definite" b ; a and
or
of "f"(x) between integral
fx I
"j)(x)dxor
\{s(x)\[s(a) "
where a
the upper limit is left undetermined "corrected" integral;
"f"(x)dxor without as
the
calculus
limits
is called
-^(x)
limits and regarded merely specified any of reversal of an operation the differential is called
"indefinite"
an
"
or
unconnected
"
integral. 12. Addition
of
a
Constant.
the differential coefficient of ^]s(x\ it is also the differential coefficient whatever of \lr(x) C is any constant + C where ; for is zero. the differential coefficient of any constant
It will be
Accordinglywe
This
constant
obvious
might
if
that
"p(x)is
write
is however
not
usuallywritten clown,
but
GENERAL
METHOD.
will be
understood
STANDARD
in
exist
to
FORMS.
21
definit all "cases of in-
integration though not expressed. of indefinite processes frequentlygive results of different
13. Different will
Idx }*/I-x2
instance is the
,
is sin'1^
what
differ by
is a
cos~%, for
"
;
reallytrue
for
,
Vl pressions. ex-
cos'1^.
"
is that
sin"1^
and
"
cos"1^
constant, for
f
that
so
form
differential coefficient of either of these Yet it is not to be inferred that sin"1^^
But
or
integration
1
dx
"7
sin ~lx
=
Vl-a2
or
,
dx=
"
cos-^
J/s/l-a;2 the
constants beingdifferent. arbitrary
14. Inverse
Notation.
verse Agreeably with the acceptednotation for the intions, and inverse HyperbolicfuncTrigonometrical we might express the equation
j5^)
or
and
it is
useful occasionally
=
^); to
employ
this
notation,
INTEGRAL
22
CALCULUS.
character which very well expresses the interrogative of the operation are we conducting.
satisfied
Laws
15. General
by
the
Integrating
Symbol \dx. (1) It will symbols that
plain from
be
the
meaning
of
the
s
but that
constant. is "j"(x) + any arbitrary l-y-(fi(x)dx
is distributive; operationof integration
(2) The for if u, v,
functions of
be any
w
x,
-T-j |u^+l^^+l^^r and
therefore
constants) (omitting
\wdx l JurZ#+|i;cfe-f =
is commutative (3) The operationof integration with regardto constants. (I'll
For
if
-j"
=
v,
and
be any
a
d
constant,we
have
du ,
so
that
(omitting any
constant
au
or
which
=
of
integration)
\avdx9
a\vdx=\avdx, establishes the theorem.
METHOD.
GENERAL
several
Integration of
-
d
"
xn+l nrfll
-
dx been
(as has Art. 7, Ex. 1)
Thus
the of
the index
For
+ l
n
already
seen
in
Art.
4
and
increased.
so
example,
/nA X
X
r =
5
Q
r
11
~"''}X x=if
;
}x
EXAMPLES. TTn'^e down
the
of integrals 1.
O
x
^"J
1
0
""}
b
a
O.
in
of any constant integration the index by unity and divide
for the
rule
is,Increase
x
obtain
we
^
+ 1
_
_
Hence
of
xn.
n
by
23
now
By differentiation of
power
FORMS.
proceed to a detailed consideration forms of functions. elementaryspecial
16. We
17.
STANDARD
"#"
^7999 #1000.
-^"Jf
c
x~^
T4=
1
"4^4-
INTEGRAL
24
18. The
Case
CALCULUS.
of x~\
It will be remembered
that x~l
or
is the
-
differ-
x
ential coefficientof
Thus
logx. fl
\-dx
logx.
=
Jx This
therefore
forms
apparent exceptionto the
an
generalrule f
^n+l
\xndx =
19. The
result, however, may be arbitraryconstant, we
deduced
Supplyingthe
/xndx
C
+
=
=
A
~
is still an
Taking the
-I-A
"
"".
+ l
n
and
limitingcase.
n+l
C+
=
a
have
n+l
where
as
constant. arbitrary
limit when
-
l=0,
+
n
takes
-
the form
logx,
[Diff.Calc. for Beginners, Art. 15.] and
as
C is
we arbitrary
infinite
portion
may
suppose
togetherwith
-
7i
it contains
that
-
another
a
tively nega-
arbitrary
~\~J.
portionA. Ltn==-i {xndx logx + A.
Thus
=
20. In we
the
same
in the
as
way
have 1
(n + V)a(ax+ b)n
=
and
of integration
+ 6) ^-log(a% "v
= "
xn
and
STANDARD
METHOD.
GENERAL
f/
therefore
IA
'
=
"
"/
y
I '"r Jax + b
printed as
a
I -r,
(n+l)a
shall
we
"
Jax
,
-\og(ax + b\ 6V
=
fFor convenience
often
dx
,
Jja2+x*
+ b'
25
(oo5+ ")n+1
7
\(ax+o)ndx
and
FORMS.
as
f
1
Jax
+ b
find
I J*Ja*+ ,-
"jdx
"
o,
etc.]
x*
EXAMPLES. Write
down
the
1. ax,
integralsof
of1, a+x,
x
"
a"x?
x,
1
x_ a+x a? x
a
2
a
'
a-\-x
3. a
+ x
a+x
21. We
may
coefficients of
a
"
bx
a-x*
next
(a-#)2" (a x)n* "
x+a
x
remark
and ["f"(x)]n+l
d
"
(a+x)2
(a
"
xY
that since the differential of
log$(x)are respectively
and
{["t"(x)]n"t"'(x)dx
have
we
=
and
The
is of great especially the integralof It may thus : be put into words use. is the differential any fractionof which the numerator is coefficient of the denominator log (denominator). second
of these
results
"
For
example,
INTEGRAL
/ /co\,xdx J
CALCULUS.
xdx
*"
=
Sill
/tan .#
dx
=
I
"
"
log sin
#,
log cos
x
X
"
J
=
.
-a?^
=
"
log sec
=
x-
cosx
EXAMPLES. Write
down
the
integralsof
(a
nex,~-, ~\~Ct
"
22. It will the that
perceivedthat
the
of operations
IntegralCalculus are of a tentative nature, and in integration ledge success depends upon a knowof the results of differentiating the simple
functions. of
be
now
standard
It is therefore forms
the practically
which list
same
and differentiation,
the
necessary is now
to learn the table
appended. It is already learnt for
that proofsof these as
results lie in
of the the righthand members differentiating results. The list will be graduallyextended supplementarylist givenlater. PRELIMINARY
TABLE
OF TO
RESULTS MEMORY.
TO
BE
several and
COMMITTED
a
INTEGRAL
28
them. as
x
The
For
it is
the
through therefore
are
Also
is obvious.
reason
increases
and
instance,x
Each
first
of these
functions
; their
quadrant
negative. help to observe
further
a
CALCULUS.
the dimensions
being supposed linear, /
a
"
i
dimensions.
There
2 d~"
tan"1-
-\-X
be
integrationmust be
be
no
"
"
a
"
is of
side. zero
X"1
prefixed
-
dx
C J
therefore
v
"
of each
efficients co-
to
the
in-
a
/
tegral. Again
not
could
"
decreases
differential
2
is of dimensions
of dimensions
(which
is of
Hence
-1.
-1.
Thus
dimensions).
zero
the
the result of
integralcould student
The
should
a
therefore
have
factor
is to be
-
in remembering difficulty
no
in which
cases
the
prefixed.
a
EXAMPLES. Write
down
the
indefinite
integrals of
the
tions following func-
:
"
'
2.
3.
*
cos2-,
coss# .
sin #,
2 4.
cot
x
+ tan
x,
cos^f
-+-^-
.
\sin^7
#e + e*
'
log sin
x
V
snr^/
^sec-1^.
\/^-i*
CHAPTER
III.
OF
SUBSTITUTION.
METHOD
The to
V
by
z
being if
Or
the
any we
formula
the
To
variable
independent
prove
change
x
of
formula
x.
be
this, it is only
necessary
to
\Vdx\
=
=F.
then
du
du =
--
dz
whence
the
changed
V=f(x),
write
u
But
be
may
F(z), by
=
function
will
Variable.
Independent
the
of
Change
25.
u
dx
dx
dz
I =
J
~rrdx
V-j-9
"-_-=
-j-.
dz
dz
write
from
x
INTEGRAL
30
Thus
to
CALCULUS.
integrate/
-dx, let tan~1#=;s.
Then
dx
1
n
and the
becomes integral
*" dz
26. In
after
formula
usingthe
choosingthe
it is usual
to
form
make
of the transformation x F(z\ of differentials, writingthe =
use
equation
j^=F'(z)dx as
the formula
Thus
be
will then
of the left hand in the
side
dx reproducedby replacing and x by F(z). by F'(z)dz,
we precedingexample,after puttingtan~1^ti=0,
write
may
*=d* -
and
I+x*
27. We is
F'(z)dz]
=
a
next
definite
The
one
l+.r
consider the between
result obtained
case
when
the
above, when
x
=
F(z) is
Let then and
if the limits for
a;
be
integration
limits. specified
a
and
b,we
have
OF
METHOD
Now and
when
when
SUBSTITUTION.
x
=
a,
z
=
F~
x
=
b,
z
=
F~
31
\a) ; \b}.
f{F(z)}=-j^,{F(z)}
Also
and whence
with rethat the result of integrating gard f{F(z)}F'(z) limits F~\a) and F~\b) is identical to z between with that of integrating f(x)with regardto x between
so
the limits Ex.1.
a
and
Evaluate
6.
/
x"z^^ and
Let
^x Ex.
2.
Let
^3=2;,and
Evaluate
/.
Put
^?=tan
\Txdx.
N/a?
therefore
cosfjxdx"
-
cos
-
J
dx=Zzdz
I -cos2. J z
/.Aos
=
/cos
2-2(^2=2
z
dz
=
2
smz
J
x^dx. 3xPdx=dz
therefore
/^2cos x*dx
;
;
llcoszdz ^smz=^ =
e, then dx=sec20d6
;
when
#
=
0,
we
have
0
when
#
=
1,
we
have
0=.T ;
=
0, 4
sin x\
CALCULUS.
INTEGRAL
32
ir
f-T^=-dx \J\+*
:.
P *?"| sec20 dB
=
fsec0 "
"4 Let
6^
=
^, then
dx
f
Evaluate
sec
0
sec
-
=
V2
1.
-
jo
dz.
=
=
0 dB
{
\ Tsech^^]. [i.e.
_x
x
e'
exdx
B tan fsec
=
sec#
{ =
Ex.4.
ir
When
when
#=0, 0=1, and
Hence
3=e.
rtan-'/V= ten-
=
2
""
tan-1
-
=
t
Ji
L
"
o
The
indefinite
is tan~V. integral
EXAMPLES. 1.
Integrate
(Put
excosex
-
cos(log x)
^=4
(Pat logx
=
4
x
2. Evaluate
\-=-,dx (Put x*=z\ "
J 1+^4
reintegrate+ fl ^ Evaluate acos#
-
-,
a^sin
I*
"
J l-f#6
(Put a*=z). v
ex + b tanh
x.
(Put ^+1=4
"
a
Q
5. Evaluate
/"
6. Evaluate
/
7. Evaluate
dx -
a*
/*"
(Put x-\=z).
"da?
J 2V^(1+^) 8. Evaluate
9. Evaluate
[
-1
dx.
/ J 2W#
dx. -
1
(Put
(Put ^=02).
#
METHOD
NOTE
HYPERBOLIC
THE
ON
SUBSTITUTION.
OF
33
FUNCTIONS.
28. Definitions. it is desirable that the purposes of integration the definitions and student shall be familiar with fundamental propertiesof the direct and inverse hyperbolicfunctions. By analogy with the exponentialvalues of the functions sine,cosine,tangent,etc.,the exponential
For
"
e-*
"
_
PTP
__
2
2 are
ex-e~x
ex+e~x _
~
e?+e-*'
written respectively cosh x,
tanh#,
29.
Elementary Properties.
We
have clearly
coth
~
C/
-,
tanh
x
=
x
=
V
X "C"JLJ.iJ.l
e~x
sinho?
etc.
-
"
t"rihx
-=cosh
^
2 sinh
x
cosh
x
=
2
"
:
^
"
"
.
-^"
=
^
with many other results analogous to the formulae of Trigonometry. E.
i. c.
=
sinh2#,
"
AM
c
common
CALCULUS.
INTEGRAL
34 30. Inverse Let
us
function
Forms. for
search
the
of
meaning
the
inverse
sinh"1^.
Put t
then
x
=
smh
y
and
"
Thus and
=
y we
shall take this
=
x"
=
log(x"
expressionwith
a
sign, positive
viz.,log (" + "v/l + #2) as sinh"1^. 31.
Similarly, puttingcosh~1x
y,
=
have
we
ty+e-y x
and
"
cosh y
==
e*y
and
ey
=
whence
y
=
and
"
JL
we
shall take this
x"*Jx*-l, log(x " *Jxl 1), "
expressionwith
a
positive sign,
viz.,
/#2
1
"
32.
Again,puttingi"nh-lx x
and
as
=
tanh
y
=
y,
"
have
-
e2y=-"
therefore
1"
whence
we
tanh
-
lx
=
x
4-log S ^ "
1"
"-
05
INTEGRAL
36
cu-l
x
whence
CALCULUS.
tan-
=
0, 2
and
tan
a;
eu-l
x
tan2 2"
=
=
^TTT
0_e-
.e-
!~4^-: ~2~ Hence
^
=
tan
~
1sinh u
=
gd u.
logtan(j+|)=gd-^
Thus
of
the inverse Gudermannian
x.
EXAMPLES. Establish the 1.
2.
results : following
"
/cosh#cfo?=sinh#.
4.
/sinh
5.
xdx
cosh
=
x.
/sm,.^o?.r
=
"
sech
x.
J cosh%
J 3.
/cosech2.rc^= -coth#.
6. (sech2A'dr=taiihtf. J
7. Writing sg results :
x
for sin
gd x, etc.,establish
"
(a) /
the
following
SUBSTITUTION.
OF
METHOD
37 ,*
36.
and
Integralsof
-4-f
X
differential coefficient of
The
loge"
-
*
.
log"
=
z
..
Jx/^2-^2 the
resemble
log
=
a?
"
--
the
38. We
,
established
put
=.
t
cosh
a
=
Hence
u
=
W^2
+ tt2
du
oj
leZu t6 =
J
"
u,
cosh
a
f =
a
u
integrate *A2-tf2. dx
=
u.
=
a
sin 0 ;
=
a
cos
have
we
du
-
"
a?
az
sinh"1-.
=
Integralsof
then
:
a
=
Let
a
the results thus
\/x2+
and
=
39. To
viz.,sin'1-*
sinh it, then
a
=
Similarly puttinga? a cosh Fa sinh u f dx I!" ri Jx/x2" a2 J " sinh /
^,
results
dx
)*Jx*+a^ dx
these
aid to the memory.
might have
f To find I
,
a
integral I
for the
analogyis an
.
a
Va2-*2 and
1 l-
^ =cosh-1-.
hyperbolicforms
inverse
that
-
a
\/x2
+
x
,
,
37. In
,
a
dx
F .,
0.
bimilarly
smh
=
"
0 d$,
is
7 laK
J
11^
=u
=
cosh~1-. a
INTEGRAL
38
CALCULUS.
{+/tf-^dx
and
=
ia sin 6
.
0+
cos
a
or
-
-^
sina
40. To
integrate
Let
cc
cf^
then
sinh z,
=
a
=
acosh0
=
cosh2z,
=
a2\ cosh2z
,
then
we
1 + sinh20
since
have
IJ^^dx
=
|a sinh
0
dz
.
a
cosh "z+ -^-
.....
.
Va2
.
,
2-smh-1x
OF
METHOD
41. To
39
integrate
Let
x
dx
then then
SUBSTITUTION.
since
cosh%
1
"
JJsif^cPdx
a
cosh z,
=
a
sinh
=
sinh20,
"
dz ;
z
a2 sinh20 dz
j
=
C62?
sinh
Ja
=
\2
z
a
.
"
^-,
-log-
or
we
0
a2!
a
"
cosh
42. If have
we
put
tan#
therefore
", and
=
__
[by Art. 40.] tan
sec
x
05
-
,
h
"
^
snce or
_
-
_
2 cos2#
+
,
J log^ n
.,
J log(tana? +
-,
"1"
sec
^),
INTEGRAL
40
43.
Integralsof
Let tan
^ 2t
z
=
;
CALCULUS.
cosec
and
x
sec
x.
differential takingthe logarithmic
1
9x
dz =
or
"
^
2
x
dz
dx
7
^dx
-
z
-;
"
"
.
z
smx
n,
2tan2 xdx"\
Icosec
Thus
In this
example let x dx
Then and
sec
Isec xdx
Hence
44. We
have
=
2*
logtan
(T+ 9)-
=
logtan
(r+s)
or
STANDARD
a
dx
,x
=1"g
JTP^l
l+Jx2 a2dx "
*x-
FORMS,
g
Jx/^+o1
I+/x24 a2 die
~
x+\/x*+a?
dx
"
""
the
now
\\/a2 x2dx
^.
+ y.
logtan
. ,
f
=
=
ADDITIONAL f
-=
logz
dy,
=
ydy
=
"
=
=
fi2 =
2G"
a
OF
METHOD
Icosec
dx
x
=
SUBSTITUTION.
logtan^.
Isecsccfo =log tanf^-
EXAMPLES.
3.
the
down
Write
of integrals
x
x
4.
5.
7.
1 8.
cosec
2#, cosec(a#+"),
^'
cos
tanV
3sin^-
x
-
"/
Find
sin ^7+ 6
1"
/cosec#cfo?=logtanby expressingcosec
10. Deduce
11.
a
-sin2^'
^j
\SQexdx by puttingsin
x=z.
x
as
CALCULUS.
INTEGRAL
42
that
Show
12.
/ /
cosh
dx sec
x
=
Integrate
13.
1
tflogtf'
when
lrx
represents
log
log
the
log
log
^7,
being
repeated
...
times. r
15.
Prove
[ST.
PETER'S
COLL.,
etc.,
1882.]
CALCULUS.
INTEGRAL
44
with
\dx integral 0'(#) lx/r(#)cfo
new
a
which
be
may
easilyintegrablethan
more
the
original
product. be put into words thus rule may Integralof the product "j"(x)\{s(x) 46.
The
Integralof 2nd Integralof [Diff.Co. of
1st function
=
-the Ex.
1.
another
cos
important
integralin
by
1st x Int. of
2nd].
possiblejxcosnxdx
with
nx.
to
which
if
connect
the factor
has been
x
if x be chosen the be done as may second integral"$(x\ i.e.unity,occurs
Thus
"
x
Integratex it is
Here
:
removed.
$(x\ placeof x.
function in
This
since in the Then
the rule
/"l.5
(xvxnxd**,*!*"?J
'
J
n
sin "
x
If
9^7
-
cosn"N
n\
n
sin
n
cos
nx
I
-
"~~~~\
/
nx
'
Unity may integration.
47. an
Thus
be taken
as
/1 logx
/logxdx"
of the factors to aid
one
dx
.
=
=x
x
logx log x
"
"
/x
"
I \dx
-(logx)dx
TION
INTEGRA
repeatedseveral f 9 / #2cos
Thus
PARTS.
45
operationof integrating by parts may
48. The
mt.
B T
be
times. #2sin
? dx
nx
f sin / 2#
nx
0
"
"
-
nx
7 dx.
-
J
J
n
n
and /
n
finally,
f J
$x^ J
Hence
=**** "
dx
nx
-*\_^COS nL
n
#2sin
Zx
nx
n
cos
I
_
~
2 sin
nx
nx
7^
T9
*
of the subsidiary into 49. If one returns integrals form this fact may be utilized to infer the the original result of the Ex.
1.
integration.
/eaxsin bx /eaxcos
and
if P therefore,
and
/eax$m we
have
and whence
aP
dx
=
"sin
bx-~\
e^cos
bx dx
=
"cos
bx+-l
e^sin bx dx ;
Q
stand
bx dx
+bQ
dx,
for respectively and
=
bx
/eaxcos
eaxsin
bx
dx,
bx,
-bP+aQ=eftxco8bx" nrct
n P=eax-
sin bx =
"
b
cos
bx
"
r?
a2+o2
/
and
w+v "
(a2 +
(bx 62)~Yeaxcos \
"
tan"1
-
aJ
).
INTEGRAL
46 The
should
we
will observe
student
that
by puttingn=
obtain
CALCUL
US.
these
results
"
the
are
same
that
I in the formulae
^)""ss""it^"^(^+"^)' [Diff.Gale, for Beginners, Art. 61, Ex. 4.] And
this
obvious.
otherwise
is
rxsm/j^,\ jg fag
game
to
as
the
increase
multiply by r
/
pnsi^
""
angle by tan"1-, the
if to
For
factor
a
differentiate
Va2 +
62 and
to
""
is the
which integration,
a
inverse and
diminish
Ex.
divide
operation,must the
angle by
Integrate\/a2
2.
"
xl
out
again
the
factor
Va2+62
tan"1-.
by
the rule of
by parts. integration
r _
,
_
J A/o2^2^=
[Note this step.] c
%
a2sin~
l-i
CL
Iv a2
-
"
J
and dividingby 2, whence, transposing
which
Ex. Here
agrees 3
with
the result of Art. 39.
Integratee*xsm2x e3xsin% cos3#
cos3^.
"
=
"
-=
-x _(2e3a:cos -
"(1
"
cos
4#)cosx
TION
INTEGRA
Hence, by Ex.
"\
"
-j=
cos
1r j
tan~
16, p. 55, Diff.Gale, for Beginners,putting
Ex.
l in the
"
^34
3\/2 ,: V
"
fx
cos(3^-^--^cosf5^-tan-1|) V 3/J 4/
J_
n=
47
^
-
[Compare
TS.
PAR
1, cos3# dx
I e^siiA
B Y
result.]
EXAMPLES.
Integrateby parts : 1. xex, x^e*,xze?)x cosh #, ^?COS2 2. ^?COS^7, ^2COS07, sin
sin
sin 2# sin 3^.
3.
x
4.
#2logtf, ^nlog^7,^n(log^)2. e^sin^costf,e*sin x cos ^ cos 2#. eaxsin^ sin qx sin r^?.
5. 6.
x
cos
^
#,
7. Calculate
x
|^sin^^,
0
0
8. Show
9.
Integrate Isin"1^^,/^sin"1^^, \
Geometrical
Let
PQ
and
Illustration.
referred to be any arc of a curve Ox, Oy, and let the coordinates of P of Q (xv y^).
Let
PN,
be
QM
area
PNMQ
the
ordinates
and
rectangular be (XQ,yQ),
PNV QM1 pointsP, Q. Then plainly
abscissae of the
But
0
that
50.
axes
/*x sin2^pc?^1,/
=
rect.
area
OQ
-
PNMQ
rect. OP
=
f
-
area
the
48
INTEGRAL
and
area
CALCULUS.
PN^M^Q
cv\
ri
Thus
Let
us
o
o
consider the
now
I x dy.
=
curve
J
to be defined
by
the
equations and and
y let
the values of t corresponding to #0, y0, and a^, 2/1 of cc and y respectively.
t0and t"be
the values We then have
ri
ri
I *0
and
and
7
,
r*i
2/"x^=l vdu=\ *"
"0
I o?c?2/=l udv**\
TION
INTEGRA
that the
so
equationabove
and thus the rule of
B Y
may
PARTS.
49 '*
be written
by partsis established integration
geometrically. 51.
Integralsof
Ia^sin Reduction
the
Form
dx, I#mcos
nx
nx
dx.
formulae
for such integrals the above as readilybe found. Denote them respectively by may Sm and Cm. Then, integrating by parts,we have at once cos
m~
nx
and(7m= Thus
and
Om= ,sin7i#
cosnx
____+m^
=
and
771(971 v___ "
, "_
Cm=
"
---
m(m
,cosnx
-l"-
----
1)
"
"
n
Thus and m
the four
when =
l
the
found, viz.,
are
n
"Sn "
E. I. C.
for integrals
^
I sin
cos
me
nxdx=
J
,
n D
cases
m
=
CALCULUS.
INTEGRAL
50
sinrac
f
xv
GT0 =
at. o\ =
I
7
cos
nx
dx
nx
dx
sinnx \"
eosnx
Icesin
t
\ cos \x J
=
,
7
,
=
x
"
"
^
~
0
=
T^CC ax
=
by
be deduced
all others
can
the above
formulae.
Integration by
for
Rule
the
of
52. Extension
of applications
successive
Parts. If
and
u
be
v
x
we
rule for
denote
with respect integrations of the extension prove the following may by parts, integration
\uvdx
where
dashes
and
x
suffixes
differentiations and to
of
functions
=
uvl
"
u'v
for
u^n~1^ is written
\uvdx
u
with
=uvl
TI
"
1
dashes; for
\u\dx,
"
Vufv^dx =u'v2 "\urfv^dx, \vtf'v2dx =u"vz
Iu'"vBdx ufv^ =
etc.
Iu(n l)Vn_1dx
=
-
=
Vuf'Vzdx,
"
"
Iu^'v^dx,
etc.
u(n~ Vvn
-
Ivf^Vndx.
CALCULUS.
INTEGRAL
52
have
we
bx dx
Ixneaxsm
=
eaxsin (bx
"
J
d")
"
eaxsin(bx
"
r2^
r
"
^^~
r3
n\
eax{P sin bx
or
Q
"
cos
where X COS
-
3-
Q=
sin 0
"
n
"
...
xn~^
xn~l
xn
30"
"
^-
sin
20
+
n(n
"
1)" ^- sin 30
"
...
Similarly
L^a*cos ix Ex.
1.
Since
we
have
dx
=
eP*{Pcos bx+Q
sin
bx}.
Integrate ix^smxdx. \e*smxdx
S^e^sinf.r
"
-^J,
f^3ea:sm^^=^32'^ea;sm('.r ^ 3^22~Vsin^ -
2" .
-
-
VVsinf ?" 4 .77
-
\
6 .
2~VsinCr- TT)
/
=etc.
Ex.
2.
Prove ^|
-rjQ ^s
^Vto^ito-^-iy^j^^ /r=n EXAMPLES. 1.
Integrate (a) femai"~lxdx.
(5) (sfaitr^xdx. (c)
(d)
/"
(e) \
Ixv"Pxdx. '(/) /"cos-1^.
TION
INTEGRA
2.
Integrate (a) [x
(")
PARTS.
B Y
dx. sm"1f
(c) /sin-1'
/^5^"'. tan
(d)
lx
-
r
/ptn
-dx.
~
53
pin tan
-
lx
dx.
(c) J
*-*
dx.
4.
../*..
Integrate (a)
I
(b) I xefsm^x (c) I cosh 5.
/log
Integrate
-
J
6.
Integrate
7.
Integrate
8.
Integrate
9.
Integrate(d)
.
...
e(suix + cosx)ax.
\x
(e) I^22*sin
dx.
(/) /cos
sin bx dx,
ax
r ..
(a)
b log"\dx. -j
sin'1^ dx.
x
/cos201og(l+tan 0)dO.
J*4"|^.
^^
TKJPOS) 1892"]
(")
[a, 1892.]
1-cos^
i
/\
T-"
10. Prove
11.
i
that j
Integrate
2." dx.
/" c?2v
/u
7
du
c?v
2dx=u
"
"
"
/(asin%
+ 26 sin
v
x
C d^u
cos
i
/ v-"dx.
+
"
x
+
c
cos2^)e*^ [a,1883.]
INTEGRAL
54
12.
where
Show
the
that
if
u
be
series within
CALCULUS.
rational
a
the
integralfunction
brackets
is
of x,
finite. necessarily
[TRIN. COLL., 1881.] 13.
and
If
Ieaxcos bxdx,
u"
that
14.
Prove
v
Ieaxsm
"
bx
+ v2) (a2+ "2)0*2 =
dx,
prove
that
era*.
that
-"
m+1 Also
m+L
that
(m+1)2
(-ir-^!?
3
^"-1 where 15.
I stands Prove
for
logx.
that
(i.)
{e^w J
+^""j)"2 leax^n-'2bxdx. J a?+ri2b'2
[BERTEAND.] 16. Evaluate
/x* log(l x^dx, -
iT5
+
277
+
3T9
+
and
deduce
that
-==9""310ge2'[a,1889.]
V.
CHAPTER
FRACTIONAL
ALGEBRAIC
RATIONAL
FRACTIONS.
PARTIAL
FRACTIONAL
ALGEBRAIC
54.
\
-
-"
of
Either Fractions.
FORMS.
of
Integration
or
^
and
-"
forms
these
-9(x"a\
-
*"2
a?
"
FORMS,
should
be
thrown
into
Thus
=___
a2
x2
_
2aj\x
"
1, =
a^
log
s-
^
Ja2"
#
a
F
1 =
-
;"
4-
+
x
a
"
"
"
2a
f
a
"
L
a
i^"l .1 coth"1
"
a
J
a
=!f(-J-+_J_Yfo 2aJ\a4-a)
a;2
1
a
"
F
a+oj ,
=
^"
2a
x/
l,i
T^l
=-tanh"1-
losr"
.
toa
"
a;
La
aj
Partial
CALCULUS,
INTEGRAL
56
(Compare the
brackets
of the results in square
forms
1 .
for
the result before tabulated
with
C
dx
!~n
1 =
"
Jcr+or
viz.,
-=
x\
)
tan'1"
-
o
a/
a
dx
Integrationof
55.
f-H
Let
6
J
a
dx
=1f.
f
J
a
c
2
AV_^2a/ 2J "
a^a
\
4a2
dx or
2
take
we
as
Thus
the former b2 is " or
if 62
"
"
=
"
/
--
;
.
tan
"
"
l"
cot
-.-
~
differ at most by constants,but expressions givencase a real form should be chosen.
These
cording ac-
4"ae, I
or
arrangement
4ac,
7_
If b2
latter
4ac.
"
coth"1"
or
any
the
or
in
56.
FORMS.
FRACTIONAL
ALGEBRAIC
RATIONAL
of of expressions Integrals
the form
57
*
px + q
be obtained
can
at
px + q
tion followingtransformapb 2a (2ax+b)
by
once
_p
the
,
~~
the
of integral
the first part being
^" log(ax2+bx+ Za and
that
c),
part beingobtained
of the second
by
the last
article.
[The beginnershould obtained.
the above
notice how
form
is
of the coefficient of the differential
It is essential that the numerator
Jirstfractionshall denominator, and
be the that
all the
#'s of the
numerator
therebyexhausted.]
are
T? '
=
57.
+ J log(^2
4*7 +
5)
-
Although the expressionpx
2
tan-1^ + 2).
+ q may
be thrown
into the form
we by inspection, might proceedthus
Let where
:"
pa?+gsX(2oaj+6)+/i, X and
/x are
constants
to be determined.
by comparingcoefficients,
giving
X
=
and
pb =
--
Then
INTEGRAL
58
CALCULUS.
EXAMPLES.
Integrate 1.
f
4.
f fo+1)^
5.
/Jfl-LZ-^p.
xdx .
2
/"
^a^/7/y.
//y"
.
/ 0^t1
c^
"
6.
58. General Fraction and Denominator. of Expressions
\2
I
J x2+?
x* + 2x+l
3.
f/v"
Rational
with
the form
A~4,
Numerator
f(x) and
where
"/"(#)
9w
rational
functions of x, can be integralalgebraic by resolution into Partial Fractions. integrated of putting such an The method expressioninto Partial Fractions has been discussed in the Differential Calculus forBeginners, Art. 66. When the numerator is of lower degree than the denominator the result are
consists of the A
And
A
sum
of several such
Ax+B
terms
as
Ax+B
and
the numerator is of as high or higher degreethan the denominator we may divide out until the numerator of the remainingfraction is of lower when
degree. The terms of the quotientcan in that be integrated and the remainingfraction at once be put into Partial Fractions as indicated above.
case
may
A
Now at
once
fraction of the form any partial into A log(x a).
"
-
integrates
-
A
Any
fraction
of the
form
-.
ix
1 r"l
"
"
^~*
^ a)
A
(x"a)r~v
into integrates
CALCULUS.
INTEGRAL
60
integralis
and the
Ex. 3.
^
/
Integrate
Put
x
\
"
the fraction becomes
Hence
Dividingout
until
=
y3 is a l+
2y
1311111
and
therefore ^2
and
the
"We
now
a?
1
is integral
Ex.4. Let
1
Integrate =
1
+y
divide
?/.
Aj/
factor of the
the fraction
Hence
=
-dx.
; then
out
by
remainder,
ALGEBRAIC
RATIONAL
is a factor of the remainder. until ty4 coefficients : detached use 2 + 3 + 3 + 1
To
shorten
the
work
(J
) 1+2+1 l+
FORMS.
FRACTIONAL
f+f+
|
i-i-i j+t+ f
i
+
-f-f-i f+"+f f +tf +if
+
A
tt-A-A 551
ll-5y-
e
Now and
11
by
Rule
-5j/-5y2
=
ll-
5(^-1) -5(^-l)2
Gale, for Beginners, 2, p. 61, of the Diff.
5#2
\(x)
1
and 3
3(^
+ 1
3
l+#
3
x
Thus I!
^.2
-L
i
ijj
^
~
+ 1) 2(^7 1)4 4(tf I)3 8(^7 1)2 (a? 1)4(^3 5
^-1) and
the
-
-
-
-
integralis plainly
1
48
A7
1
1
+ 1
6
(2a?-l)-3 ^2-
61
we
INTEGRAL
62
CALCULUS.
EXAMPLES. 1.
Integratewith
v11*'
w'
2
^+ (iii.)
regard to
x
the
followingexpressions:
\V11V
T~\*
a)-1^ + b)~\
~f~
(viii.)
\7
?T7
"a*--") "
(iv) -/
"
"^
2. Evaluate
3.
Integrate (i) W
4.
dx f J (^2+a^2+62y
(iii) "'
f
J
Integrate
(xd* (i.) v ; J^+^2
.
+ l
do?.
[' (iii.) v
-
J^+l
(iv.)f
cto. Aa?2"t1
J^4-^2+l
(v.) r (vi.) /"(^-
\'
5.
FORMS.
FRACTIONAL
ALGEBRAIC
RATIONAL
Integrate /.
xdx
v
.
.
dx
x
dx
(vii.)
(^"T4)-
^
(iiL"
^"**"
W
("") (x\ VA*/
6.
~(~"
lX-^-4)' (x*+ i\/ i
Integrate J~.
~3,J~*
(VI.) -7 x""
d^t?
\
f
~t
\
"/
/T
j
j ^
\o/i
~t
""
:
\
V*-'"^*^
o\'
(viii.)
''
J
+ iy (#-l)2(#2
/Vtan~"^(9 and P\/c
7. Evaluate
8. Obtain
the value
c
o
cos
9.
x
dx
Investigate
10. Show
*
^2
that
r. fa "o
_f^
+
1)3
63
CALCULUS.
INTEGRAL
64
that
Prove
11.
dx
[+*
_2?r 2?r
a
b
+
~~
J
(x*
"ax+
"
"2X^2
bx
+
b'2)
V3
ab(d"
[COLLEGES 7,
Show
12.
be can
and
that
expressed
of
the
the
infinite
1891,]
series
sum
in
the
form
that
hence prove
[OXFORD,
1887.]
VI.
CHAPTER
STANDARD
SUNDRY
doc
f
60,i.
I.
When
a
where
of
Integration
Case
METHODS,
R
a
Positive.
is
positive
we
write
may
this
a
we
may
dx I
Q p
____.^=._.=_..___^_^==i
aJ 7/
a
If
I __
.
_
__
x/"J
bz-ac
"\2
as
as
arrange dx
If
integral
dx
If
which
ax2+2bx+c.
=
-y=
,
+-
according of
form
62 is
as
the
greater
integral +
ax
E.
T.
~
b
1
C.
as
ax .
or
"2 is
"
"
or
smh "7^
Va
ac
"
and
ac,
the
real
(Art. 36)
*
Vo2
according
than
is therefore
*
^
less
.,
,
cosh
=
or
, ~
b
+
T 1 ,
x/ac
, "
62
INTEGRAL
66
In either
the
case
CALCULUS.
be written
integralmay
in the
form logarithmic
^
the constant
"
*Ja*Jax2+ 2bx+c),
b+
log(ax+
~T=
_
62
logv
T=
*J a Also since
cosh
lz
~
beingomitted,
ac
~
=
sinh
l\/z2
~
1
"
,
sinh
and 1
,
cosh
=
I
and
.
.
1
-1" "
=
"
7--
\/ac
V^
b2
"
l\/z2 + 1
~
"7^
,
cosh
,
\/aR
,
sinh
"=.
+ b
ax
, "
cosh 1
= .
,
sinh -7=.
=
+ b
ax
, l
~
lz
~
"
x
x/aJi
T 1 -
-?
7
\/a
\/ac
"
b2
forms therefore may be taken when a is positive b2 is greateror less than ac respectively,
which and
Case
61.
If in the
write
a=
"
II.
Negative.
a
integral A.
Then r
ZJ
or
"7=:
sin
'
"
/
)*Jax2+2bx+c
1
or
dx
f
our
.
integral may dx
a
be
negative
be written
INTEGRAL
68
Ex.
dx
(
Integrate
2.
CALCULUS.
J This
integralmay
written
be
dx
I
therefore
and
is
"
sin"1-^^"
=
.
\/2 which
also be
may
\/41
expressed as -^cos
-F="
*/41
V2
EXAMPLES. 1.
Integrate
{--^" JV^
2.
62.
dx
dx -,
J
A/2"Ja + Zbx+cx*dx
3.
Integrate
4.
Integrate /\/a +
\
of
Functions
J
+ 2a? + 3
/"
Integrate
dx
{
2"#"
the
3.* -2#2
(c positive).
cyPdx
(cpositive).
-.
Form
-"=====
x/a^2+26^+c be
integratedby
which, or
may
first
be done
as
putting Ax+B
in Art.
by equating coefficients Ax+B ex/
;
we
into the
may
iorm
57, either by inspection obtain
of integral
The
METHODS.
STANDARD
SUNDRY
69
the first fraction is
A
that
and
of the second
been
has
discussed
in Articles
60, 61.
EXAMPLES.
Integrate
2.37+ 3
-
x+b
POWERS 63. Sine Index.
PRODUCTS
AND
Cosine
or
SINES
OF
Positive
with
Any odd positivepower of immediately thus : integrated
sine
a
COSINES.
AND
or
Odd
Integral
cosine
"
To
integrate Isin2n+1# dx, let .'.
smxdx=
cos
x
=
c,
"dc,
Hence
fsin^+^cfo ((I-c2) dc =
-
__
can
be
CALCULUS.
INTEGRAL
70
Similarly, puttingsince we
therefore
s, and
=
cosxdx=ds,
have
Icos*n+lx dx
(1
=
s2)nds
"
L_
nn
64. Product
sin^
of form
cos?#, p
1 \7l.
/
I
'
""
q odd.
or
product of the form method admits of immediate by the same integration either p or q is a positive odd integer, whenever whatever Similarly,any
the other be.
.example,to integrate/sin5#
For
therefore
/cos%
Hence
sin5^?dx
" "
/^
sin5^ cos3# dx
cos9^? 9cos7^;
of tan
in terms For p + q
put tan =
"
x x
=
or
cot
we
"T"'
proceedthus
"
negative
a
even
of immediate
integer,the integration
x.
t,and therefore sec2^ dx
=
dt}and let
2n, n beingintegral.Thus =
)n
|
~
^
8a5
Irftan^+6a5
4-n~*-(j ,
"
p + i
:
-
admits
sin*tocos% expression
J^f~
I sin^(l sin2x)d (sinx)
=
is
and
"
~5~'
p+^
cos#=c,
/c4(l c2)2dc cos5^7
65. When
dx, put
dc.
sin xdx=
-
cos4#
__
2
I-
4--
ldt
if Similarly,
METHODS.
STANDARD
SUNDRY
put
we
cot
x
c,
=
then
"
71 cosec2^ dx
"
dc,
and
\"DPxco"xdx=
a
result the
-
the former
as
same
the
arrangedin
posite op-
order. Ex.
Integratef?^"fo?.
1.
J
This may
sura
be written -
and
/
the result is therefore
It may
also be
CcosPx sPx =
J sin6^?
Ex.
1
f
I
,
-r-^-dx
the result
integratedin -
/T
o
terms
2
.
of tan
an
^
thus
:
"
tan~5#
\,,
"
x
x=-
"
-
"
-
J taii6
being the
same
as
before,
2.
/sec" (9cosec"
66. Use
of
0 d"9
=
ftan~*0rftan0"" -f tan~%=
-
f cot*0.
Multiple Angles.
of a sine or cosine,or Any positive integral power of sines and integral any product of positive powers in means cosines,can be expressedby trigonometrical series of sines or cosines of multiplesof the angle, a and then each term be integratedat once; for may f
\cos
J
siunx nx
7 dx
=
and
"
n
ifJ
cosnx
7
sin
nx
dx=
--.
n
INTEGRAL
72
f /"-t-UU"^^
r
/^^2^^r^,_
J}x x.
i
Ex.
2,
1
/ cos
.
c/Lx =:
x
cos%
Ex. 3.
CALCULUS.
Sin 2.27
X
/
.
dx=
/"
/
_
j/(|+ J cos
=
%x + J sin
"
67. It has is odd
alreadybeen
the second
more
we
Ex.
4.
Let
cos
shown
that when
the index
is necessary,
nor
for the
x
+
c
sin
x
2
cos
=y
2t sin
x ^
in
of
case
x
x
;
then w
=
=
y
"
-,
2i sin
nx
"
yn
"
"
yn
Thus
2
cos
8^-
16
cos
6# + 56 cos4o?-
112
The be
of
sin^cos?#, where
y
=
in
"
odd.
are
q
sm
sin
Integrate I8m9xdx. x
thus
different form. will therefore discussing
result
now
are
=
-
value especial
neither p
g1^sin Ax.
/(1 sin2.2?)a? sin x
=
presents the
method
2# +
example / cos3# dx
which
J cos 4dc)dx
2# +
transformation
such
no
""**
"4"
cos
2^+70.
l(cos 8x
sin8.??
Thus
=
METHODS.
STANDARD
SUNDRY
8
-
6# + 28
cos
4#
cos
73 56
-
35),
2# +
cos
2
f
andj
"
K
irsiii8#
j
\$n$xdx="
2i L_
J
Ex.
2
=
and
cos
8#
8
"
.
"
"
-
Kcsin2^?+35#
-56
.
"
Q*
"1
"
4
o
2^
,
_J
; then
6^+8
cos
7J
sin6^cos2^="
OQsin4#
+ 28
"
/ sin6a7cos2^ o?^.
.#+t sin ,#=y
cos
Osin6^
-8"
"
8
Integrate
5.
Put
-
-
'8# + 4
"cos
8
4#+
cos
cos
6^
"
2^;
cos
4
cos
kx
10,
"
4
"
cos
2^7 +5
V,
whence
68. NOTE.
It is convenient for such examples to remember that the several Coefficients may sets of Binomial be quickly in the reproduced followingscheme :" 1 1
1 121
1331 14641 1
5
10
10
5
1
1
6
15
20
15
6
1
1
7
21
35
35
21
7
1
1
8
28
56
70
56
28
8
1
etc., each
number
being formed above
the 7th
row
0+1
we
=
1,
it and have 1 +
5=6,
at
the
once
as
the
precedingone.
5 + 10
=
15,
10 +
sum
of the Thus in
10=20,
one
mediatel im-
forming
etc.;
CALGUL
INTEGRAL
74
US.
/
and
in
occurring above and
such
multiplying out
onlyneed
we
all the work
(1 t)Q
coefficients of
(1-/)6(1 + 0
6 + 15
-
(1
will discover
the several coefficients
the
are
"
6 +
1,
5 + 9-5
+
-
1+4
+ 6 + 4 +
1+3
+
coefficients here
1-0=1,
O2
"
+
law
same
by
c,
were
6-4
4-1=3,
=
1,
required.
thus
:
"
etc.
4-6=-2,
2,
EXAMPLES. 1.
Integrate odd
doing those with 2. Integrate
3.
indices in two
ir
/
4. Evaluate
ir
r"
sin^ctr,
*0
5.
ways.
Integrate ft
Integrate sin
6. Show
r
/ cos5^?c?^, ^0
0
si 2.# cos2.r,
that
/sin x 7. Show /"
/
sin 2.# sin 3#
dx=-"\
cos
2#
-
""$cos
4# +
^\ cos
6#.
that N
f
7
.
(i.)I sm
wia?
cos
w^
a^7
=
"
cos(m+?iV v ! "
"
"
as
forming per-
the required,
2-2-3-1, + 2 +
1,
c+c?,etc.
1,
formed
are
6,6 +
+
1,
cZ2+cfa3+... by l + ",
+
(1+04(1
the coefficients
are
row
4+4-4
of this
reason
a
a,
1+2-1-4-1 the last
5-
of a+fo multiplication
Similarlyif the coefficients in work appearingwould be
The
-
+ t)2 (1 t)G(l
-
the actual
and
15
20+
-
9-
1-5+
are
student
The
in which
coefficients of
1
+ (;*/
)
-
4-10 are + 1)* 1-4+ 4+ + t)G(l figuresbeing formed accordingto the
of
before.
1
are
-
coefficients of row
the
1\6/ "
appearingwill be
coefficients of
each
(y
product as
a
^
cos(m 72-V y )", "
"
.
INTEGRAL
76 and
CALCULUS.
generally s"5
^3
2w+2ajdx where
c
70. Odd cosecant
By
=
can
c
-
cot
=
"c"
-
WC"-
-
"
have
we
.
.
.
x.
of
positiveintegral powers thus : be integrated
differentiation
/"2n-fl ~
at
a
secant
or
"
once
d "
and
(n + l)cosecn+2" n
cosec7lo?
"
=
x cosecna -7-(cot
"
doc
whence 1 ) secw+2^jdx
(n +
tan
=
x
secn#
+
cosecnx
+
^
Ise
and + (ti
1 ) cose.cn+2xdx =~coix f
Thus
as
x
Icosec
and we
/
sec
infer at
may
dx
c?x
#
cosecn#
n
dx
I
\
=
+ ^V logtanf^
=
logtan^ ,
of integrals
the
once
sec3#,sec5#,sec7cc, .
.
.
by successively puttingn
; =
cosec3aj, cosec5#,etc., I, 3, 5, etc.,in the above
formulae. Thus
/sec3#
dx
^ logtan f
J tan
x
sec
/sec5^?da7= J tan
#
sec3^+ 1 /sec3^
J tan
x
sec3#+f tan
=
=
x
+
^7 sec
etc.
^V
+
-
x
+
f logtanf
-+-
),
SUNDRY
formulae
71. Such
METHODS.
STANDARD
A
as
student
77
called
are
"
"
REDUCTION
with
will meet
others
formulae, and
the
in
postpone till that chapter the of the integrationof such an expression been have as except for such cases
Chapter
VII.
consideration sin^cos^
as
many
We
alreadyconsidered. 72. Since is
a
a
of or
a
cosine
a
cosine
73. may
that
POWER
OF
tannx
dx
=
=
TANGENT
tann
and we
#
cfc
Itan%c dx
tangent
a
~
=
COTANGENT.
we
have
/tan3#cfo?=
cotangent
or
2x(sec?x l)dx "
f,
"
logsec
,
tan?l-2#cfe.
J
1 #,
(sec2# l)dx "
=
tan
integratetan3#, tan4^, tan5#,etc.
may Thus
==
a
able
now
"
n
Itan
are
Itann-2a3c?tan# ltanw-
-^
since
of
power
negative power
OR
idM.n~lx
And
we
positive
a
cosecant.
of Any integralpower be readilyintegrated.
For
negative
a
cosecant
or
sine, and
integralpositiveor
of a sine, cosine) secant, or
INTEGRAL
is
sine
or
secant
a
or
it will appear
cosecant
integrateany
to
of
negative power
power secant
of
positivepower
/tan x(sec2x-l)dx
x
"
x,
78
INTEGRAL
CALCULUS.
[
f
2
3
By continuingthis
process
shall
we
evidentlyobtain
_
2?i-l
2^-3 # + (-l)nff, (-l)w~1tan
+
tan^^
tan-+^=^^
and
t
_
Similarly Icoinx dx
cotn
=
~
2
cot71-1^ =
"
~ --
r-
J
71"1
whilst
icot^^aj
=
cot2^ ^
=
and and
therefore
we
f |COtn-2#CfcE,
logsin x. I(cosec2^ 1 )dx
may
"
thus
admits
"
integralpower of a tangent of immediate integration. f
Integrationof
We
may
write
a
+ b
x
"
#
or
cotangent
dx
etc. \a+bcosx, cos
x
;
etc.
any
74.
cot
integrate
cot3#3 cot4#, cot5^, Hence
=
as
j, s2|sin2| -
METHODS.
STANDARD
SUNDRY
79
(a
or
(fa
2
Thus
-AgU-j
=
"
6
"(1)
or
CASE
I,
If
a
"
b this becomes
tani a-
6
/a+6
,
tan
or
?| 2J-
Since
we
may
write
this
as
"
b,
1-
1
^ =5- COS
"
g
+ b
2
INTEGRAL
80
CALCULUS.
1 or
COS"
"
7
*-"
+ bcosx
CASE
II.
If
a
in the form b,writingthe integral
"
(K
dian.~ ,
(2)
in
placeof
the form
UjiAj
A J fti _1_ + bcosx f*na
A b
by
case
Art. 54
J.
"
sv*
in this
have
(1)we
n a
"
IT. 1 Ib +
"
17 a
"
Ib+
Vf^ v6 +
,
a
x
V6^~tan2 a
+
a
"
\/b
"
a
tan
a
tan
"=
"log '
.
rjr"
.
\J~b+
b
v
"
^ "
By
Art. 33 this may
be written
tanh~:
/62-a2
or,
we
since
2 tanh
~
lz
=
cosh
~
1
1
may
stillfurther
02'
"
exhibit the result b 1
"
a.
b+
a
1
1
"
^"-
b+
or
nX
tan2^
-L-^
:cosh~3
as
a
2
STANDARD
SUNDRY
We
METHODS.
81
therefore have
b,
"
x
i.e.
dx bcosx
a+
er "
or
=
Jl?-
cosh-
a+bcosx of the real is used.
but These forms are all equivalent, forms is to be chosen when the formula
75. The
integralof
b.
"
one
may
"
-r
a+
b
cos
x
+
c sin
be im-
x
mediatelydeduced, for b cosas-fc
smx
\/b2-{-c2cos(x tan~V ), b/
=
"
\
the proper form of the integral at can be written down in each of the cases a greateror
therefore
and once
less than
^/5*+c*. dx
Ex. 13 4- 3
cos
x
in H- 4 sin
=f [ J
x
dx --
1
/132 -
52
13 + 5
_
12 1 or
$. I. C.
-i/2
.'T
"
a\
(where
:-a) coe(# a) -
tana
=
^)
CALCULUS.
INTEGRAL
82
dx
f 76. The
integral I
.
7
Ja + 6sm# ,
be
may
--
"
easily deduced
by putting
dx
f
then
"
j". o sin
Ja + therefore
and the
by
Ja +
x
its value
dv
"^o
-
cos
,
y
be written
may
down
in both
a^b.
cases
Of
f I-
=B
"
course
be
it may
also independently investigated
first writinga + b sin
x
as
+ 26 sin | + sin2|j a(cos2| |, cos
+ 26 tan cos2^(
or
a
The
then integral
-
+
a
J. tan2^
becomes
2
and
two
cases
77, The
arise
as
integral
before.
x
I" ,
may
,"
treated. dx
f
dx
be
similarly
CALCULUS.
INTEGRAL
84
certain
that,with
4. Prove
limitations
on
the values
of the
involved
constants
/"%==L=
-
J J(a-x)(x-(3) P .-(3
____
and
/
integrate
(x
v
a)(/3 x)dx. "
"
a
5.
Integrate ,.
JSC
("'") /"""
dx
C
v
"}
.
.r
\ (1""
Ou27 cos
a
+
f
}3(l-s
(v-)
.
U11'-'
J
r
"^%
r
\
,
'
/
*
\
^'''^
COS.T'
-
2^2 r
4- cos
and
+ cos
a
x
(viii-) prove dk
Integrate(i.)f(ii.)/
^ V
C?Jt' ____
a(^
-
f
b)+
V
6(^ a) -
"**
(iii.) J
7.
Integrate 7f I
8.
Integrate
f- ^ -
J sm^
9.
Integrate
-
.
+ sm2^
fcos201ogcos^+shl fa. COS0-S1TL0
J
10. Interate
1+cosx
.77
J"2siii2"9 + 62cos2^'
o
6.
+ sin
^^
(vii.) cos
a?
11.
Integrate
12.
Integrate /
_dx.
+ sin
x
sec^
Integrate /
dx.
"
J 1+
14.
x
sm
J VI
13,
METHODS.
STANDARD
SUNDRY
x
cosec
Integrate /-"f^fl_f_. J
v
b tan2#
+
a
fVr^
15. Evaluate
"^'
"
/ 1 +
x
sin
o
16.
Integrate [****"****"",. J logtan ^7
17.
Integrate
.
Vsin 2(9
fcot0-3cot30
18.
IntegrateJ
19.
Integrate / J Wo?
20.
Integrate /"7
~
J (x si
21.
Integrate f-^
22.
Integrate fA/_ V
'
*
CQS
~
cos
+ cos (9(1
1
sin
^ 6y
6")
+ cos 6")(2
f
23.
24.
25.
Integrate Integrate f^ (sin0 + Integrate f J
.
" .
"
^?
2
"
sin
a?
"mg-coeg cos
85
CALCULUS.
INTEGRAL
sin"1
I
Integrate
26.
dx. -
.
l+x2
J
27.
?
Integrate J
\"
( sin^, 28.
{^^dx, and
that prove
J
'
sin
sin"r
[*"-X-dx,
Integrate 2#
J
sin
3^
J
sin
4^7
5 +
[THIN.
COLL.,
1892,]
CHAPTER
VII.
REDUCTION
FORMULAE.
FORMULAE.
REDUCTION
Many
functions
immediately
reducible
79.
forms, In
and
some
cases,
connected
of
whose
by
another
to
the
For
be
itself
form
in
the
Such
methods
terms
be
formula itself
any
rate
of
integral
that
shown
of
terms
may
with
standard
obtainable. be
linearly
the
integral
be
may
easier
either
mediately im-
integrate
to
of
(a2 + #2)^fe
J(a2+ #2)^fe,and
which J(a2 + ar)^cfe,
I(a2+ #2)^cfcmay
connecting
Reduction
80.
it will
in
at
the
not
are
function.
original
instance
integrals
which
or
of
directly
not
are
algebraic
some
integrals
other
or
sucft
however,
expression,
expressed
one
integrals
integrable than
whose
occur
being
be
relations
algebraical
this
can
latter
standard
a
inferred.
are
called
Formulae.
The
student
have
will
already
realise been
that
used.
several
For
reduction instance
the
CALCULUS.
INTEGRAL
88
parts of Chapter IV., and It is proposed to consider and such formulae more fullyin the presentchapter, of some for the reproduction to give a ready method of the more important, of Integration method by the formulae A of Art. 70.
where
81. On the integrationof xm-lX* for anything of the form a+bxn. In
several
the
cases
be
can integration
X
stands
performed
directly. I. If p be
the positive integer,
a
expandsinto
binomial
finite series, and each term
a
in
is integrable.
/v"
Next and
s
p fractional
suppose
=
-,
r
and
8
beingintegers
positive. 777/
II. Consider
the
Let
X bnxn~ldx
.'.
when
case
=
=
a
"
+ bxn
is =
szs~ldz zs~l
\x
r-
J
and
when
is
"
bn)
a
this expressionis integer, positive
directlyintegrable by expanding each term. integrating III. When
"
positive integer.
zs,
f
and
a
is
a
the
binomial
and
the expression negativeinteger,
(zs-a)~n+'
FORMULAE.
REDUCTION
may
be
may
then
partialfractions,and the integration proceededwith (Art.58).
into
put be TD
If
IV.
r
H
"
is
"
proceedthus
may
89
we integerpositiveor negative,
an
: "
rn -
, _
rn
m-\ and
by
is
either
III. this is integrable when
II. and
cases
positive or
a
substitution
777
or negative. integral, positive,
S
n
Three
thereforeadmit of integration by simple substitution.
cases or
(1) p (2)
positiveinteger.
a
"
an
integer.
[-p
an
777
(3)
Ex.
Here
"
integer.
Integrate (^(c
1.
m=6,
n
=
3, and
=an
"
integer.
n
Let so
that
Then
the
"
r
\-- is
"
"
negative integer by the That is, the expression is
a
b + ax~n=-zs.
integrablewhen
-
--
%x*dx=
integralbecomes
2zdz.
mediately im-
INTEGRAL
90
Ex.
Integrate/ x*(a?+ x^dx.
2.
Here
CALCULUS.
m
=
", n
3, p=b
=
and
is
+p
"
integerc
an
'
n
is / integral
The Let then
-3-. XT
the
and
becomes integral 9 *
might be put
which
6, the
process of will be avoided effected (Art.70). =
sec
formulae
82. Reduction + 6xH
Leta with
xm
then
^C;
=
of the
any
fractions. If,however, z be put partial tions puttingthe expressioninto partialfracand the final integration may be quickly
into
\xm~lX^dxcan
to according
Let indices
P
=
n
-
-
1^
the
and
m+n
lX?+ldx,
-
Xm+n
rule following
"X+1JTya+1 where
of x
connected
"
\xm-n-IXPdx, xm
be
six integrals : following \XP
-
\xm~\a
for
X
A
-
~
1
: "
and
JUL are
in the respectively
two
the smaller
expressions dP
whose
are integrals
Find
-p.
arrange Re-
linear functionof the expressions and to be connected. are Integrate, integrals this
whose
to be connected.
the connection
as
a
is complete.
CALCULUS.
INTEGRAL
92
Integrating, P=(n
/"( 1) I(x2+a?fdx-na?
+
and n+I
Putting?i
=
5 and
ft
3,
=
(( J
and Then
3i
6.4
Ex.
3.
Calculate
the value
of
[^x^-^ax-x^dx,
shall endeavour
positive integer. We
m
being
to connect
with fxm~l*J%ax-x*dxy l%m"JZax-x'*dx
(xm~\2a-x*fdx. with (xmJf^(Za-x?dx
i.e.
Let
P=^m+1?(2a-^)1r accordingto
the
rule,then
Hence
(m
+
xfdx 2) fxm+^(2a -
-
xm^(2a^ -
+
(2m + l)a fxm~\2axfdx -
a
FORMULAE.
REDUCTION
a
93
.
xm*J?Ltix o
ra
Jo
+ 2
+ 2
m
o
/la, .
_
xm*j%ax
x*dX) and
-
be
m
2ra-l
2m
to find
Now
1
+
IQ or
+ 1
5
"
dx
=
sin (
a
sin 0.
^l^ax
Also
when
#=0,
when
#
"
Hence
70=
Hence
/
x^
"
=
fVsinW^-
2a,
0).
we
have
$=0,
we
have
O
T(l -
1)...3
-m-
(m+2)(m
+
m '
3
"cos
a
d
an
3
x^dx, put
x=a(\ Then
.
etc.
_ "
'
"*4
m
fj^ax
I
3/ J.m-z
2m-3
-1 '
m
2
m
'
"mT2
a
-
.
--
-
m
-3
2m
2m-l .
positive integer,
a
l)...3
cos
+2?r_ 2
=
TT.
20)rf0
(2m + l)! m!(m + 2)!
EXAMPLES,
Apply reduction 1.
/ J
2.
the
rule formulae
stated
(when
in Art. 82 to X=a + bxn): "
obtain
the
following
INTEGRAL
94
3.
CALCULUS
(,-^ J
4.
( J ^P
=xm^ (a"*-*X*dx
.
{x"+n
mm]
J 6.
-
/
.
Integrateout 7. Obtain m
l, m=2,
=
of
integralsof
the m
3, and
=
are integration
values
when
cases
the limits
2a.
formulae
similar rule may
be
for
sin^a? co"x
givenfor
Isiupx
for
the
"
their numerical
0 and
83. Reduction A
/xm^(^Laxx^dx for
cosqx
a
dx. formula
reduction
dx,
j
be expression may six integrals : following This
with
connected
of the
any
"
I sin^
"
\sinpx
Isin^ by
the
Put smaller two
-
dx,
\si
^x dx,
\si
2# cos?#
cos9'
~
rule. following sinX+1#cosAA+1" P where =
indices of since and
whose expressions
Find
sin^+2^ cos^
2x cosv+2x dx,
-T-,
and
cos#
are integrals
rearrange
as
a
-
and
X
*x
dx,
^
are
in respectively
the to be connected.
linear
functionof the
ax
whose expressions
are integrals
the
to be connected.
FORMULAE.
REDUCTION
the connection
Integrateand Ex.
Connect
the
95
is effected.
integrals
/" P=s
Let
=(p (p
"
=
[Note the
"
l)smp~2xcosg^(l sin2#) (q-f l)si cos9# I )sin^~2# (p -h ^)sin^ cosPx "
"
last two
lines of rearrangement
*
Hence
.
P=
(p
/sin.^
I ) /siii^~2^7 cos9^
-
8m
cos%^
=
dx
(p
-
*~^X cosq+l*
-
+
where
or
q
is
q)Isi
P + qJ
in the
integral fsin*4?cos?#e"i? with 1
.
/
smp+2x
cosq.v dx.
2.
/siii^
3.
I smpx
4.
/*sin^-
5.
/ sin^
+
in
the
case
integerthe complete immediately[Arts.64, 67].
EXAMPLES. the
functionof
odd
an
integrationcan be effected The present method is useful integers. q are both even
Connect
linear
however, that
remembered,
either p
a
+"zi (*
p + q
It will be
as
sin^~2^7 cos%],
and
sin^cos^
.
"
cos"~23? dx.
2# cos"~2# dx.
case
where
p and
INTEGRAL
96
6. Prove
CALCULUS.
cousin-1*
fsin^^
that
J
n
this formula
Employ 7. Establish
f^,.
"-I
_
to
formula
a
J
n
integratesin%, sin6#,sin8^.
of reduction
/cosw#
for
dx,
Integratesin4
8.
calculate the
84. To
integrals
V
71
f 2"
.
I smn#?
5^n =
and
aa?
fl I
(7n =
J
J
0
Isinn^cdx
Connect Let P
=
0
Isinn
with
sinn~3#cos;E
~
2x dx.
the rule; then
accordingto
dP sunx
"
"
"
"
dx =
(n
l )sinn2x ~
"
"
smn~lxcosx
f lsinn^a^= J
n
.
.*.
x
J,
we
=
sin" -^
71"1
be
even
cos
^
"
---
71
71
n
vanishes
"
x
=
this Ti-l
71"2
3 ^
71
"
5 a
---
*
^
7i
"
Ti-3
"~'ii^V"g
5
""w-
4
to
ultimatelycomes
when
0, and
have
=
if
r
2, when
less than
integernot
.
-\smn~zxdx. J
"
n
since
Hence
If
"
, ---
3
Iff
4
2j J
?"
is
an
also when
FORMULAE.
REDUCTION
7i-3
Ti-1
3
1
TT
that is n-2
n
If
n
be odd
1
"
il/
Ufl
"
-L
^^
ll/
^^
Q
A,
O
TJ
9 Z"
" "
.
"
"r"
71
and
422'
similarly get
we
f-j
.v
'"
n
2
"
I sin xdx
since
97
=
*
*
\
*
rl w U**/j
dm w E*-1-11 w
"S I
5
-|
.
I
"P
3J
cos
"
/" 2"
I
r
x
1
=
o
7i-3
Ti-1 we
have
$n
=
1
In
a
4
2
-
similar way
"
2'"
5* 3* be
it may
that
seen
I"cosnxclxhas o
the precisely
in each value as the above integral This may be shown too from odd, n even. case, n other considerations. These formulae are useful to write down quickly of the above form. any integral same
/""".""",, "^|."
[The student should notice that these are easilyby beginning with the denominator. of natural numbers ordinarysequence Thus the first of these examplesis (10 under
with
most
We then have the written backwards.
is
(3 under
odd,
in
2) and
factor
forming such no
factor
-
35 E. I. C.
down
9)x (8 under 7)x (6 under 5),etc.,
stoppingat (2 under 1),and writinga first denominator
written
G
is
-.
a
But
when
sequence
written.]
the
it terminates
INTEGRAL
98
CALCULUS.
r"
85.
To
investigate
formula
a
2si
for 0
integralbe
Let this
denoted
by f(p,q) ;
then
since
tanP^^ J
p + qJ
p + q q be
have, if p and
we
less than
GASP:
and positiveintegers,
p
2
I.
If "" 6e
ei"e7i
2m, and
=
# afeo
(2m-l)(2m-3)
even
=
3)...l
m
"/v
^/^
"
9""
\
/(O,2")
/i
7/1
^i
"
1
2ft
2n,
=
)
/(
2rv
andj
be not
"
3
1
v
2
TT
=
0
Thus
CASE
II.
If p
6e
=2m,
6^67^
2m~1
q odd
-/(2m-2, 2^-1)
/(2m, 2^-1)=
' -
and
and
=
=277 etc.
"
1,
INTEGRAL
100
These
CALCULUS.
will
relations
n-{- 1 is either
T(n + 1) where
2k
being For
1
+
'
positive integer.
a
instance,
T(6)
=5F(5)= =
This not
5 .
4F(4)
5.4.3.2.ir(l)
V-) =F(f
do
to
an
2
k
sufficientlydefine integer or of the form
found
be
)= S
.
=
.
=
is
called
propose
to
enter
.
3r(3)
2F(2)
5 .4.3.
=
5!
PXiHf
function
4
5
I- fr(f )= i .
.
a
Gamma
into
.
|
.
f
.
f r(" )
function, but
its
properties
we
further
here. The
products
1.3.5...
2n-I
2.4.6
2u ...
TT
which
in
occur
the
foregoing
I
of
cases
sin^0
cos?0
o
be
may ^
so
expressed
at
once
2n-3
^(2n+l\_2n-l 1
\~~2~)
in
of
terms
this
lr/l\
2n-5 '
2
2
2~
2
that /y
2
and
sothat
Hence
in
Case
function.
I.
7T
V2/'
d9
FORMULAE.
REDUCTION
In Case
101
II.
In Case III.
we
evidentlyhave
the
result.
same
In Case IV.
It will be noticed therefore have the same result,viz.,
that in every
case
we
7T
f and
the
that the ^
sum
This
of the is
?9 1
+ l
and
#4-1
^"
the
convenient of the above quicklyintegrals a
the denominator
occurringin
+1
very
is
in the numerator.
formula
for
evaluating
form.
IT
Thus
rs f \in"6" cos80 dO
=
-*
.,
__f f j-V^TTj" f f i|^/?T "
'
"
"
*
"
_
5?T
2.7-.6.5.4.3.2.f~~215'
CALCULUS.
INTEGRAL
102
87. The student been pointed out
should,however, observe (asit has
that when either p or q previously), both of them odd the or are expression integers, without reduction sinP$cos?# is directly a integrable formula For
at all.
instance,
[sin^(l-sin26'Xsm6'=^7 (sin66"cosW6" 79 J =
J
and
Similarly, -2 0 cos26"(l cos2"9+cos46")dcos "i
COS3"9,QCOS5"9COS76n"
"Jr4-"-*^*-
+2"
-
3
But
when
p
and
are
q
if the
or integralrequired,
other than
0 and
both
^,
must
we
.jo,!
and limits of even
either
use
the indefinite be integration the reduction
tt
formula
of Art. 83
or
proceedas EXAMPLES.
Write
down
the values
of
s
in Art. 67.
FORMULAE.
REDUCTION
the formulae
prove
(1}
I sm2mOcos2n6d0
(2)
I?L_?.-_. 2
="
J
-Em+n
f sin
J 4.
103
-Bm+n-l
Write
down
the indefinite
0 dO, fsitfO
of integrals
fsitfO cos50 dO,
cos3"9dO, fsitfO
cos
fsin70 cos20 d09
cos4^ dO. fsiu60
Evaluate rT
rf
/"
sin5^cos2^^.
/ sm26
/ sin4#"w,
__"
J
J
0
0
0
/3" 7T
7T
6.
7T
/-^
/"
/"
J
"T
J
'
0
7. Deduce
the formulae
of Art.
84
/
for
(
sm x
dx from
r("ii)r(z"l) y 7 V
result
27
V
"
f
of Art. 86.
EXAMPLES. 1. Prove
that
(a) I cos2w"" ^ "/
(b)
=
-1 tan "/" + cos-nc/"
M
^^t
\
-
~
\ fcos2
2iii/ J
the
INTEGRAL
104
a formula Investigate
2.
when
of reduction
by
7.
=
of reduction
for
that
._J_ 271+2
2
27i+ 4
tegrati in-
completethe
[ST.JOHN'S COLL., GAME., 1881.]
show integral
of this
means
to applicable
and positiveintegers,
a formula Investigate
3.
and
and are n if ra=5, 7i
m
CALCULUS.
271+ 6
2.4
2.4.6
adinf
271 + 8
2. 4. 6. ..27i
~~3. 5.7...27i + l' Sum
also the series
1
1
1
1.3.5
I m
271+1
1.3
1
I
2
2^ + 3
1 _1_
I t
,
2.4
"
27i + 5
2.4.6
.
,
\f
OjCll ITlrT
271 + 7
[MATH. TRIPOS, 1879.] that
4. Prove
2n+l
/
(rf
*-, 6. Find
reduction
formulae
prove
for
(a) x"(a+ bx)P*dx,
(y)
J
and
*^*"+a")"*r, (S) /2p+l /*"(*"
obtain the value of
7. Find
a
reduction
and positive integer,
-
.
formula
for
[COLLEGES"CAMR]
Ieaxcosnx dx, where
n
is
a
evaluate
[OXFORD, 1889.]
/#wsin from
Deduce
dx
x
the latter
a
1 05
for
of reduction
formulae
8. Find
FORMULAE,
UCTION
RED
/eaxsinnx
and
formula
dx.
of reduction
for
Jcos
8in"*"fe.
a*
[COL1KGES
%
189o.]
Tt
9. If
rT / si
un=
o
that
prove
and
^-l-
deduce
un=
-^^-+^
-"
2n+1 In
--3),
rc-
rC-f1-/ /b --
-
--
sv
+
\ f
""
"(^" l)(n-2)
n("" 1)
(2ro-lX2ft-3)...3 TT '
8'
[MATH. TRIPOS, 1878.] 10.
Show
that
1 V
/
1 \
'
wi"
2/3
n"2fifJ")
[TEIN. COLL., CAMB., 1889.] 11. Prove
that 1 2.4.6 ...2m _1.3.5 ...(2w-l) TT ~2.4.6...2m 4~3.5.7...(2m+ l) 2*
7
*
'
1
12. Find
of reduction
formula
a
for
f-~=L ^
3. 5. 7.
..(27i+l)l
where
a1} a2,
13.
Show
the binomial
"
Show
that
1
coefficients.
[ST.JOHN'S, 1886.]
that
/cos
2TO
are
...
v'^
=
mx
r+-
cosm#
dx sn^
mm
4-
"
sn^
, ~"
~~
~T72 where
m
is
an
integer.
[COLLEGES
a,
1885.]
14. Show
m
CALCULUS.
INTEGRAL
106 that
being a positive integer.
[OXFORD, 1889.]
that if
15. Prove
Im,n=
sin
I cosm#
(m + n)Im)n=
cosmx
"
da:,
nx
cos
93
92
/
[if ^-l^(2+i+i 1
If
16.
,1
prove and
that
/m?
cos
cos%^7
r
,
show
I cosw#
=
w
/m
w=
"
d f cosmx\
^
^
m2-^2
-
-
)+
m(m"l)T v m*-n2
./^m-2, n,
that
cosm^7 sin
Hence
-+^J- [BERTRAM]
cEr\oos9u;/
/I prove
9"i\
+
dx,
nx
"I
-
-
m/m_i?n_i?
+
nx
that
^m '
w=
1
--
"
+
m
find the value
?i
cosm#
dx\
wm_i
"
M_I.
+ n
m
(when
/If
7^^;
m
is
a
sin 2mx
of integer) positive
dx.
[7,1887.]
o
18. Prove
that
' r2 / cosnx
cos
nx
dx=
J 19. If
m
+
n
be even,
/
prove
IT "
T
2n+1"
[BKRTRAND.]
that
co -
m-n. i
2
1
. -
2
[COLLEGES, 1882.]
CALCULUS.
INTEGRAL
108
Find
28.
reduction
a
formula
for
the
integral
xmdx '
(log#)n' Find
29.
reduction
a
[OXFORD,
1889.]
for
formula
xmdx
C
[ft 1891.]
that
Prove
30.
if
X=a
/nZ""=[ST. reduction
Find
31.
formulae
\CLJ
(Q\
JOHN'S,
1889.]
for
I tann
x
dx.
dx
f
.
J (a+"cos#
32.
Establish
parts,
u
and and
the v
following
being suffixes
functions
+
formula of
integrations
x"
csin#)n
for and with
double
integration denoting
dashes
respect
to
x
:
by
entiation differ-
"
/ /u
+
(-
l)n-1nu^n-1hn+i
+
(-
l)"n I uMvn+idx
+
(- l)n
{dx (u^vndx. [a, 1888.]
VIII.
CHAPTER
f INTEGRALS
The
88.
of
dx
\^
FORM
OF
integration
EXAMPLES.
AND
METHODS
MISCELLANEOUS
expressions
."
of
the
form
dx
be
can
If
be
The
Let
X
and
II.
X
linear,
III.
X
quadratic,Y
and
performed,
89.
Y
I.
X
CASE
best
in
effected
readily
I.
be but
X
are
Y
both
for
cases
which
functions
linear
of
x.
quadratic. linear.
Y
both the
all
the
quadratic is
process
and
substitution
-fe
Y is
both
:
"
dx
more
linear.
integration troublesome.
can
INTEGRAL
110
CALCULUS.
Putting cdx
,
we
,
nave
=
at/.
=
-(y2
,_
and
+ b
ax
e)+ b,
"
C
and
becomes
/
21
"
the standard
forms
Jy Ex.
which, being
"
^
jay2
7 + bc .
"
ae
2_^ "A
is
2,
,
one
of
immediatelyintegrable.
f
Integrate /=
"
J (xLet
then
Thus
y+lj
y-l
90. The
same
viz., substitution, *Jy=y will suffice l(fi( T\f] 'IT
for
the
rational F
of integration
I
-^^
when
functionof x, integralalgebraic
eacfe linear.
are
Ex. .
^(cc)is
Integrate /==
f
J^-
Writing "/^+ 2=7/, we
have
%dy
and
.r
=
?/2 2, -
and
X
any
and
METHODS
MISCELLANEOUS
so
-"L ="="
that
EXAMPLES.
AND
+
24/-32/
Ill
+ 16
division).
(by common Thus
91. CASE The
II.
proper
linear,F quadratic.
X
substitution is : "
X=\
Put
y Let
Putting
ax
+ b
=
-
,
t/ we
differentiation, have, by logarithmic
dy
adx ax
and
ex2 +
ex
+ b
y
+/=
6Y + ft) -2((+/ a\ a2\/ -
-
Hence
the
has integral /=
form which
has been
been
-
/
/
reduced
: "
alreadydiscussed.
to the
known
INTEGRAL
112
Ex.
f
Integrate /=
Let
./
then
#+l=y-i,
CALCULUS.
and
__=_
#+1 /=
y
__
i+i-2
i+%-y2
V
JI*
92, It will
now
that any
appear
expressionof
the
f
form
J( be
being any integrated, "j)(x) For by algebraicfunction of x. can
"b(x) we
can
express
-
,
.
in
rational
integral division
common
"
,,
the
torm
Af
Axn+Bxn~l+
...
+Z
being the quotientand
We thus have reduced the process of a number of terms of the class integration
remainder.
M
the
to the
Eaf
and
of the class
one
f
M -dx.
and latter has been discussed in the last article, of the former class may be obtained by the integrals
The
reduction
formula
^(^_^/)4_2r-l TO
2r
r-lf
e v
c
'
v
r
c
METHODS
MISCELLANEOUS
where
F(r)stands
f
for I J
The
J
By
dx.
,
\Jcx*+ ex+f
(x+l)*/x2+l
*2+3*+5
division
exercise.
an
as
f ^2 + 3^+5-^.
Integrate /=
Ex.
113
xr
this is left
proofof
EXAMPLES.
AND
*=x
+ 2 +
-
Now
and
to
x
integrate /
we
-
and
put #+!=_
Thus
93. CASE The
III.
proper
X
quadratic,F linear.
substitution is : "
+/Y=y.
Put
T /=
Let
f
dx
I
-
-
"
J (ax2+ bx + c)V
Putting
*Jex+f=y, edx
and ax24-"^ +
E. I. C.
c
7
reduces to the form
j
"c
+/
get
INTEGRAL
114
Now
e)Ay* be thrown
can
"
-j
4
p
.
dy
-f
I becomes
and
CALCULUS.
2
partialfractions
into
n
i
as
and
each
94. It
fraction
is
is also
evident
the
that
substitution
same
of of expressions integration
for the
be made
may
by foregoingrules. integrable
the
form
*(")
f
dx"
_
J
"p(x)is
where when
^/ex+ fis put equal \
the
__
c)\/ex+f rational,integraland bx +
(ax2+
form
to yt y"
/,/2n-2_i
/j/2nj_ \
"^
algebraic ;
for
reduces
to
"
"
0 2
7
_i_ "L
which '
by divisi"n'
-
and
the
rules
for
partialfractions,may
as
and
Ex.
each
term
is at
Integrate
once
integrable.
/=
Putting \/^+ l=y,
we
have
-7====2e?y,and v^
+
1
_
2
be
expressed
INTEGRAL
116
CALCULUS.
/becomes
Thus
(a2 "
W
Also that
so
and
/ reduces
Thus
If
a
6,we
"
further
may
to
arrange
/
as
i
/"
V^TP
,
-
Ex.
2.
.
-,
,
Integrate/=
1
.
/ -
J (2x2 \/3^2-2^ 2x2 -2^+1) -2^ __
dy
3^-1
2#
+ 1
"
1
_
~~
values yj2and yf of ;/2 and minimum are given 2 by x \ and # 0, and are respectively and 1, so that for real be not greater than 2 and not less than 1. values of #, ?/2must The
=
maximum
=
MISCELLANEOUS
2^2
/ becomes
/2#"
r("g3-2ar+l )(2a72 J x(x-l)
Now
EXAMPLES.
t-tfm^l-g
and
_
AND
yi-f=^-y"='
Now
Thus
METHODS
-
2.37+1
~
1
Thus
'=/(-,== 2
=
cosh"1?/ +
2
cos"1-^-
,
1
/3^2-
N/2 \2^-
EXAMPLES.
Integrate 1.
4.
2.
5.
3.
6.
_4 "
"
117
CALCULUS.
INTEGRAL
118
96. Fractions
ina;+CCOSa;. CT+f + Pisin^ + CfiOSX
of form
al This fraction
into the form
be thrown
can
A
"
(Oj+ ^sinx A, B, G
where
A +
and
each
97.
+ c1cos x)
chosen
so
c-fos x) that
-Bc^ + Cb^b, is then integrable.
Ca^a, term
the expression Similarly a
be
may
x + (ax+ b^siu
constants
are
~
1
1
(a
+ b sin x-\- c
cos
x
arrangedas
+6^+0
.)"+
cos
_|
-
x + (Oj+ 6xsin
third fractions may
the first and reduction formula and
c-[Cosx)n
be reduced
while [Ex. 25, Ch. VII.], is immediatelyintegrable.
98. Similar a
+ b sinh
ai + 99.
remarks x
+
c
a
x
by
+ b sinh
x
+
c
cosh
x
#' (ax+ 61sinh # + qcosh x)n' x + Cjcosh frisinh
Some
Special Forms.
It is easy to show
that sin
a?
Isin^r sin
"
c\
a .
"
'sin(a 6)sin(a c) "
"
a
the second
fractions of the form
apply to
cosh
~
a),
AND
METHODS
MISCELLANEOUS
EXAMPLES.
119
sin2,^ and
"
"
7
"
"
r"
-.
-.
r
-,
T\".
a)sin(^ 6)sm(# "
c)
"
1
sin2a 1/ ^^i E51111 1
""
sin
f I
whence
-r",
r)\"nY\( rt (-t/ U lollll
/
,
" ^^
__ ^^
Iv
/^ I/ I
"
"
M
-^"7
/
/Y
tv^i
" ^^
\
.
.
-
-
7
,
r
r
/,
.
. v
.
a), lo^sm(o3 v 6 "
c)
" "
"
sin2^ dx
f
,
ollll
c?a?
#
.
Ssina 6)sm(a sm(a (a 6)sm(a and
/7* cC"
QTTn
r
-
--
-5-7
r-^-p
,
J sin(^ a)sm(x "
.
/ x o)sm(x
"
c)
"
sin2a
S
-
"
sin
;
(a
"
1\
integrate any
tan
:
/
"
7r
.
2
"
*
shown
has
a
"
"
6)sm(a c)
"
generallyHermite of the expression
100. More
x -
-
how
to
form
8,cos 9) /(sin sin($ a1)sin(0a2) sin(0 an)' f(x,y) is any homogeneousfunction of _
_
"
"
"
.
where Ti
"
.
.
#, y
of
1 dimensions.
For
by
f(t,1) (* Oj)(t a2) (t a^ -
-
-
.
.
.
_
_
-
-
-
.
.
.
,
x
^" ax
which
fractions partial 1) /(a,, (ax egCoj a3) (ax "") of
ordinaryrules
the
may
(a2 ^Xag "
a3)
"
...
(a2 an) ^" "
be written
_
__
^r((ar a1)(ar a2) "
"
...
(the factor of the above
a2
(ar" aw) ^
being omitted coefficient).
ar
"
*
ar
Proc.
Lond.
Math.
"
ar
in the denominator
Soc.,1872.
CALCULUS.
INTEGRAL
120
Putting
"
tan$, a1
=
tana1, a2
=
tana2, etc.,this
=
becomes
theorem
/(sin0, cos 6) sin($ a1)sin(0a2) sin(0 an) "
"
"
.
.
.
/(sinar, cos ar) r=isin(ar OL) sin(ar an) sin($ "
"
"
...
ar)
Thus
W: /(sin0,
9)
cos
/(sinor, ~
"
"
^ism(ar
%)
_logtan^ 2
"
^ "
a,)
cos
"
7
./
x
7
"
.
.
.
sm(ar
\
"
J-'-'ii t/ctj-j.
;-:
.
an)
EXAMPLES.
Integrate sm
cos
^
^^7
cos
cos cos
cos
0
O.
2# ^
3#
cos
2a
"
cos
a
"
cos
3a
"
cos
a
"
sn
sin
GENERAL 101. There
are
^7
"
"
"
cos
a
sin 2a sn
a
D.
.
x
x
sin 2^7
K
"
cos
-
4
#(sin2,#sin2a)' "
PROPOSITIONS.
certain
on general propositions
almost which are integration definition of integration or meaning. Thus
self evident from the from the geometrical
f(j)(x)dx= J
102. I
for each is equal to \^(^) \HC0 if "f"(x) he the differential coefficient of \fs(x).The result beingultimately ~~
*
See Hobson's
Trigonometry,page
111.
independentof z
121
immaterial whether x or plainly process of obtainingthe indefinite
it is
x
in the
is used
EXAMPLES.
AND
METHODS
MISCELLANEOUS
integral. /""
1 "p(x)dx "f)(x)dx + (j
103. II.
=
c
a
a
For if
/""
pc
of "p(x) integral the left side is \{s(b) the rightside is \^(c) is the same thing. illustrate this fact geometrically. us be the indefinite \[s(x)
"
and which Let
Let dinates
the
NJP^
N^^
be 2/ N^P^ be cc
drawn
curve
the above
Then
0(#0"anc^
=
a,
=
x
=
c,
x
equationexpresses
fact that Area
+
["j)(x)dx [$(x)dx.
104. III.
=
"
b
a
For
with
the
same
the left
and
area
the
notation hand
righthand
before
as
side is side
\/r(6)T// is [ "
"
=
^e^ ^ne or" b respectively.
the obvious
CALCULUS.
INTEGRAL
122
f"/"(x)dxf0(a x)dx
105. IV.
=
-
0
0
For if
we
put
have
we
and
if
x
=
a
dx
=
"
x
=
a,
y,
"
dy, y
0,
=
I "p(x)dx= I (f"(ay)dy
Hence
"
"
o
a
=
(by in.) fV"-2/X2/ o
=
I "{"(ax)dx (by I). "
o
in
this expresses Geometrically estimatingthe area 00' QP
the
fact that, the y and x
obvious
between
O'
Fig.9.
axes,
like take as
our
O'Q,and a curve originat 0',O'Q as
ordinate
an
our
direction positive
PQ, we may if we our F-axis,and O'X
of the X-axis.
INTEGRAL
124 since
Thus
CALCULUS.
sin"^^
/
dx
smnx
=
sm"(7r #), -
/ sinw^7dx
2
0
and
;
0
since
cos2n+1#
and
cos2n#
=
=
cos2n+1(?rx\ cos2n(7rx\ "
"
"
rir
+ 137(jfo7 COS2'l 0,
I
=
*o
ft
r
/ cos2w# dx
and
=
'0
We To
0 and
propositioninto words, thus : of the form smnxdx at equal intervals "
is to add
TT
cos2n^7dx.
0
may put such a add up all terms
between
%\
up
all such
from
terms
0
to
"
and
For the second quadrant sines are merely repetitions order. of the first quadrant sines in the reverse Or geometrically, the curve about $mnx the ordinate being symmetrical y
to double.
=
#
0
^,the
=
and
whole
between
area
0 and
TT
is double
between
that
|.
Similar
geometricalillustrations
108. VII.
If
will
apply to
other
cases.
""ti /"net
\
pa
"j)(x)dx=n\ "j
it is clear that it "j"(x), of the part consists of an infinite series of repetitions the ordinates OP0 (x 0) and JV^Pj lying between bounded (x a} and the areas by the successive ordinates and portionsof the curve, the corresponding the #-axis are all equal.
For, drawing the
curve
y
=
=
=
Thus
and
f "{"(x)dx=r'(t"(x)dx= f )dx /"a
j"wa I
I
71 1 "p(x)dx. ^"(x)aa; =
I
=
etc.
METHODS
MISCELLANEOUS
AND
EXAMPLES.
125
Thus, for instance, f2"
"
o
xdx=%
sin
F
\
J
J
"
"",
j
"
sm
T
A
=4
\276u?
-
/
Bin
J
"n
2n-3
A%n-I
7 #aa?""*4
""
I
IT
2
2
...-"
2ra- 2
2?^
-.
O
Fig. 10.
have
109. We
INTEGRALS.
DEFINITE
ELEMENTARY
SOME
the
whenever
that
seen
be performed,the can l^"(#)cfe integration
definite
the
In
many
can integral "j)(x)dx
cases,
however,
the
at
value
once
of
indefinite value
of
be inferred. the
definite
definit integralcan be inferred without performingthe inwhen it cannot and be even integration, performed. We propose to give a few elementaryillustrations. Ex.
1.
Evaluate
/=
=
{*( J
Writing we
and
have
vers~
-
a
=
TT
"
A
a
CALCULUS.
INTEGRAL
126
Hence
1=
Hence
/=
~3/2f(TTf\2ay
-
| o
Putting and
we
?/=
obtain
a(l-cos0),
f'smn+10d0
/=?an+1
7ra
=
...
accordingas
n
is
or
even
n+i
down
to
? or
l
3
22
E,
odd. ir
Ex.
2.
/ logsin #
/=
Evaluate
ofo?,
0
Let
#=--#, 2
dx
then /=
and
dy
" "
;
/ logcos y dy
"
Hence
/
#
c"
rf
rl 2/=
/ logcos
=
logsm^"ir+ / logcos
xdx
jo
\
log /I (log /f
sin
^
#
cos
c?^
o
sin 2%
"
log
o IT
r"
"j
Io88inte"fo-i
0
Put
2x=z, o?^7
then then
=
/
^dz ;
Iogsin2^^=^/logsin zdz"
I f
AND
METHODS
MISCELLANEOUS
/-^2i log 2,
27=
Thus
EXAMPLES.
/=|logl. log /?
Ex.
3.
-\
r~%
\ log cos
sin xdx"
cfo?
#
=
-.
2t
J
2i
/ -^
1=
Evaluate
log
-
o
Expanding
the
have
logarithm,we
6
If
re
/=
have
Hence
x=l
put
we
we
/
"
^-dy
"
J
/
=
"^'a?^=
/
also have e
"y, dx.
"
" "
.
6
\-x
"o
Ex.
4.
Evaluate
1
-I
^=tan(9,
Put
.-.
1=1
log(tan0+cot0X0 o
/2(log
sin ^ 4-
log cos $)6"
o IT
-
2
/logsin
0 dO
=
Tr
log 2.
127
INTEGRAL
128
110.
CALCULUS.
under
Differentiation
Integral Sign.
an
to be "p(x, Suppose the function to be integrated c) which is of independent x. containinga quantityc Suppose also that the limits a and b of the integration and of are finite quantities, independent c.
Then
will
r"
-
J0(a?, a,
For
let
u
=
f6 \ "f"(x, c)dx. a
Then
u
+ Su
=
f 0(o3,
"
which, by Taylor'stheorem,
And be
if z, say, be the
greatestvalue
of which
capable,
in the limit when vanishes and in the limit Thus diminished. ^ "
=
"
a
'
'dx.
Sc is
indefinitely
The
111. contain
c
AND
METHODS
MISCELLANEOUS
in
case
EXAMPLES.
the limits a and beyond the scope of the
which
is somewhat
129
b
also
present
volume. be used to deduce many proposition may has been performed. when one integrations This
112. new
since
Thus
L=
f --
=dx
=
(x+cyJx-a
J
-* Vc
tan-1 +
\l2=2(c+ *
a
times n have, by differentiating
we
c
+
with
^
(^+ c)(^ a)
0),
regard to
times with regard to n Also, differentiating
/-
a"
a
a,
we
c,
obtain
2
"
this differentiating Similarly, c,
we
obtain o?^?
r
IJ
latter p
"2^+1
(^+c)^+1(^-a) 2
EXAMPLES. 1. Obtain
the
: integrals following "
f (i.)f(i+*)-V*"fo. (v.)J J
(ii.)An-^-xi+ J
ar)-*^
J
*?)"*"". r#-1(2-3a?+ -(vii. (iii.)
E. T. C.
I
(vi.)
times
with
regard to
INTEGRA
130
2.
Integrate (i.) (a2+
L
62
-
CA LCUL
US.
^2)v/(a2 ^2)(^2 62)' -
-
[ST.JOHN'S, 1888.] 1
a2)^^+^
(x2 +
|UL* the values /-
sin
J (cosx /"
of
f
^
+
x
dx
a)V(cos x 4- cos fi)(cos+
cos
cos
x
y
J cotfx + a\Jcos(a;+B)coa(a: + 'v} a)\/cos(^ + ^)cos(^ + y} 4.
Prove
constants
)
cLx
I
\
[TRINITY,1888.]
6"Vacos2^ + 67iii2"9+V
sin 3. Find
I~STJ"HN'S" 1889.]
that,with involved,
certain
limitations
+ Zbx + (x-p)(ax?
the
on
d,L
C% 1890-]
"
\^ Olll
"
.
(-ap2-2bp-cft
cy*
of the
values
(x p)(b2 acf -
-
[TRINITY,188G ] 5. Prove
\(cQ$x}ndx may
that
be
the series
expressedby j\r ^.v3 "
pf p etu,
_L -"
-r
...
n-
being the coefficients of the
ND N2J Nft and n having any .
.
.
real value
expansion(1+ a) 2
,
or negative. positive
[SMITH'S PRIZE, 1876.] 6. Evaluate
the
: followingdefinite integrals "
(i\
W
fl
J
l
^2
/a(a2
+
/y.2 ^'
[ST.JOHN'S, 1888.]
^72)2
o
/"""
\
^UL'
f
** x
dx
Jo(l+tfX2+^)(3
+
7. Prove
8. Show
that
#)
[OXFORD, 1888.]
f -
that
[0x^,1888.]
CALCULUS.
INTEGRAL
132
15. Evaluate
dx
(i.)f
[I.C. S., 1887.]
o
[I.C. S.,1891.] 16. Prove
f^an^^
(i.)
J
37+
sec
J a2 a
cos
[POISSON.]
4
#
cos2.2
"
beingsupposedgreater than unity. 17. Prove
[OXFORD, 1890.]
f 1-2S"fc? g
(i.)
=
-
o
18. Prove
that =
a
"
z2)
"a3
"5
""
+
" "
-
"
-
ctf+
...
3.5.7
3.5
3
[OXFORD, 1889.] 19. Prove
that
IT
2r~% /
/-7A "^"
beingsupposed " 20.
Prove
_T
.
2 1 1
1 2 o
"
Q2
1
1
1*3-4
i
o2
2 "
^"
"
PL2 " x,6
i
1.
that
[MATH. TRIPOS, 1878.] 21. Prove
that 1
1.1
''*
22.
$(x)dx= -,"),\*a
If
[A 1888.]
F $(x)dx.
-
0
[TRIN. HALL, etc.,1886.] 23. Prove
*""6^ Jf^C~^W
that
c-a?)
^-6)
provided
b
remains
finite when
x
vanishes.
[ST.JOHN'S, 1883.]
that
Prove
24.
AND
METHODS
MISCELLANEOUS
EXAMPLES.
133
and illus/""{"##) ra$(x)dx= + ""(2a-a?)}cfci7, *
the theorem
trate
25. If
and
that
show
f(x)=f(a+x\
illustrate
26.
geometrically.
geometrically.
that
Show
\
q-pj
q-p
q-p
value of the 27. Determine by integrationthe limiting of the following series when n is indefinitely great :
sums
"
'
n
/""
+ I
+ 2
n
+ 3
(iiL)_J_+
+
n
ri
[a,1884.] n
n
n
n
x
n
*
*_+
.-+
\/2?i-l2 \/4?i-22 \/6rc-32
+
-J*/2ri*-"n? [CLARE, etc.,1882.]
(iv-) "
n
sin2/" + sin2K + sin2fc +... +sin2K" !-,K beinsr 2 J 2n 2n 2n (.
i
"
"
"
integer. 28. Show
n
an
[ST.JOHN'S, 1886.] that the limit when
n
is increased
2*
n2
3n
'2n
of indefinitely
[COLLEGES, 1892.] 29. Show
that the limit when
n
is infinite of i
i
is
Apply
/*"+*.
e^a this result to find the limit of
-('+ [CLARE, etc.,1886.]
INTEGRAL
134
Find
30.
the
CALCUL
(n\}n/n
of
value
limiting
US.
when
n
is
infinite. .
Find
31.
of
the
of
sum
value
limiting
the the
n
when
n
is
infinite
of
the
Tith
part
quantities
n+1
n '
+
2
n
3
+
n+n ~
T~J
Ti~J
"V
n
and
show
Napierian 32.
show
the
of
product the
that
If that
na
it
is
same
the
to
limiting
quantities
as
of
value 3e
:
the
8, where
logarithms. is
the
always
limiting
?ith e
is
of
the
base
[OXFOKD,
equal value
unity
to
of
the
and
n
is
the
root
indefinitely
of
1886.]
great,
product
[OXFORD.
1883.]
IX.
CHAPTER
/
ETC.
EECTIFICATION, In
113.
the
of
course
the
four
next
chapters we
of obtaining foregoingmethod limit of a summation by applicationof the process of integrationto the problems of findingthe bounded by such lengths of curved lines,the areas of solids of and volumes lines, finding surfaces to illustrate the
propose the
revolution, etc. Rules
114. As idea
for the
Tracing
shall in many cases of the shape of the we
order
of
have curve
the
a
Curve.
to form under
limits
some
rough
discussion,in of
integration, author's refer the student to the we larger may Treatise on the Differential Calculus, Chapter XII. for a full discussion of the rules of procedure. The followingrules, however, are transcribed for to
properly assign
,
convenience
of
will reference, and suffice for present requirements:"
115. 1. A curve.
I
For
glance
Cartesian will
in
most
cases
Equations.
suffice to
detect
symmetry
in
a
CALCULUS.
INTEGRAL
136
(a) If
Thus
of y occur, the curve is symmetrical powers larly Simiwith respectto the axis of x. for symmetry about the ^/-axis. odd
no
4"ax is
y2
=
symmetricalabout
the cc-axis.
(6) If
of both x and y which all the powers be even, the curve is symmetrical about axes, e.g.,the ellipse
occur
both
^ y*_ a2+62~ (c)Again,if on changingthe signsof x and y, the there remains unchanged, equationof the curve is symmetry in oppositequadrants,e.g., the hyperbolaxy a2,or the cubic x3+y3 3ax. If the curve be not symmetricalwith regard to tion either axis,consider whether any obvious transforma=
=
of coordinates 2. Notice
could make
whether
the
curve
it
so.
passes
through the
the coordinate the pointswhere it crosses coordinates present axes, or, in fact any pointswhose themselves as obviously the equationto the satisfying
origin ; also
curve.
3. Find
those parallel to the asymptotes; first, axes ; next, the obliqueones. 4. If the curve pass through the originequate to of lowest degree. These will terms the terms zero givethe tangentor tangentsat the origin. 5. Find
the
y^; and
where
dx'
find where finite;i.e.,
the
it vanishes
or
becomes
or tangent is parallel
in-
pendicul per-
to the #-axis. 6. If
of the variables,say y, in terms of the other, x, it will be in the solution, frequentlyfound that radicals occur and that the range of admissible values of x which we
can
givereal values ofloopsupon a
solve
the
equation for
one
for y is therebylimited. The existence is frequently detected thus. curve
ETC.
RECTIFICATION, 7. Sometimes reduced to the
when simplified
equationis much polarform. the
Polar
II. For
116.
It is advisable
137
Curves.
to follow
such
some
: following 1. If possible, form a table of r and 9 which satisfythe
routine
as
the
"
of 9, such
as
6
=
0,
values corresponding for chosen
curve
"^, "","f
etc.
JP
negativevalues
9 leaves the
equationunaltered, e.g.,in
a(l
=
"
both
of 9.
initial line.
r
Consider
o
there be This will be so when
2. Examine
values
,
O
and positive
of
whether
symmetry a change
about
the of signof
the carclioide
cos$).
obvious from the equation of the curve confined that the values of r or 9 are between certain limits. If such exist they should be ascertained,e.g.,if r asijm9, it is clear that r must lie in magnitude between the limits 0 and a, and the 3. It will
be frequently
=
curve
lie wholly within
4. Examine
rectilinear
or
the circle r a. has any the curve =
whether
asymptotes,
circular. RECTIFICATION.
a
117. The process of between two curve
findingthe lengthof an arc of fication. specified pointsis called recti-
Any formula expressingthe differential coefficient of s proved in the differential calculus gives rise at once by integrationto a formula in the integral calculus for findings. add a list of the most We common. (The references are to the author's Diff. Gale, for Beginners.} 118.
values
In each of the
the limits of integration are independentvariable correspondingto case
the
INTEGRAL
138
the two pointswhich is sought.
Formula
in
"he Diff. Calc.
CALCULUS.
the
terminate
Formula
in the Int. Calc.
whose
arc
Reference.
Observations.
For
P. 98.
length
Cartesian
Equa
tions of form
For Cartesian Equa tions of form
P. 98.
*=/(*/)"
ds _
dt
l(dx\*Idy
P. 103.
For Polar of form
Equation
P. 103.
For Polar of form
Equation
P. 100.
For
M(di)+(Tt
is
rdr
Pp. 103,
For
case
when
curvi
given as use
when
Peda
Equation is given
dr
105.
ds
P. 148.
For
use
when
Tan
Pola gential Equation is given
119. We Ex.
1.
add illustrative
Find
extendingfrom y="
,
yi="
the
length of
the vertex and ,
to
examples: "
the
one
the limits
arc
of the
extremityof are
#=0
and
parabolaxL=kay the latus-rectum. x"%a.
Hence
Ex. p
Find
4.
and
the
between
rsina
=
CALCULUS.
INTEGRAL
140
length of the the points at
arc
f 2- -^r
=
J Ex.
Find
5.
whose
s
where
^
of the
arc
In
\^2are the values respectively.
case
involute
of
at the
beginning
of
closed
a
be is
end
oval, the originbeing within that
the
length of
I pd\fs,for
the
the
that
exceeds
by
the limits
the
portion
perimeter of of its
"
taken.
are
ellipseof
an
of
length that
a
small
circle
tricity eccen-
having
the
[7, 1889.]
area.
p2 ^
and
"
Show
Here
circle,
Curve.
given by
disappearswhen
e
^
observed
\~dj)~~]
Ex.
a
formula
contour
-jy
of
Closed
for
it may
curve,
arc
any
=
using the
whole
=
is the
p
), c^cos2^+ 62sin2^ a2(l e2sin2^ =
angle =
which
makes
p
-
with
the
major
axis.
.A a(I i^sin2^-e*sui*\ls. -
-
\
Hence
of the
are
=
Formula
in the
Hence
vectores
=r
and
120.
where
radii
VV2-r2sin2a
length of
the
equation isp
Here
same
equiangularspiral
the
which
r2.
Here
the
of the
arc
2i
o
/
I \-^-\\f}(very approximately) ,=4a{|-lA
ETC.
RECTIFICATION,
(r)of
radius
The
circle of the
a
is
area
same
141
given by
vZ^ab^a^l-erf, /
its circumference
and
-e2
"
*.
"
4
32
circle
ellipsecircumf
Circumf.
-
).
e4
"
...
\
.
\
o
I
1 27ra(
=
/
( \lo
=
"
.
\irae* =
-
2ira
""
. _
3'2
t"4
/
4
o '
of circle], far as [circ.
= "
involvinge4.
terms
as
64
EXAMPLES.
by integrationthe length of the arc a2,interceptedbetween the pointswhere
of the
1. Find =
2. Show
from
that
the
in
the vertex
catenary y
(where #=0) =
c
smh
cosh
the
-
length of
arc
point is given by
to any .
s
c
=
cos
x"a
circle and a
x
i
-.
c
3. In
v
that
the
the
where
evolute of length of the the
it meets
4. Show
that the
viz.,4(# 2a)3 27a#2,show parabola, from its curve cusp (# 2a) to the point parabolais 2a(3v3 1). a
=
-
=
"
lengthof
the
of the
arc
cycloid,
.r=a(0+ sin 0) ^ a(l-cos"9)J y =
between 5. Show
the
pointsfor
that in the
0=0
which
and
for epicycloid
y=(a
6)sin0
+
0=2^,
is
s
=
which
b sin
-
=
26
a
5
measured beiijg
from
the
point at
measured
"=--, from
a
0=7rb/a.
222
n
When
which
show
that
cusp which
4r+y*"a*j
lies
on
the
and
s3 oc y-axis,
that x*.
if
5
be
INTEGRAL
142 6. Show
CALCULUS.
that in the ellipse # be expressedas
may
7. Find
the
(i.)r
length of =
r (ii.)
=
of the
arc
any
cos
a
=
$, j/
6sin^, the perimeter
=
curves
acos0.
r (iii.)
a6.
aem0.
(iv.)r
asin2-.
=
=
2t
Apply
8.
whose
s=_"
the formula
equationis r=a(l
9. Two
radii vectores
-f cos
be
found
that the
lengthof
in finite terms
the
curve
initial line
is act, where
of the
Find
an
in the
arc
of the
cases
when
length of the arc between (c2 a2)p2=c2(r2 a2).
the
"
that
the circular
is the
a
"
two
Times.']
yn=xm+n
curve
or
"
*m
curve
12. Find
; prove
[ASPARAGUS, Educ.
integer. 11.
cardioide
[TRINITY,1888.}
of the
OP, OQ
drawn
10. Show
the rectify
0).
equallyinclined to length of the interceptedarc of the anglePOQ. measure are
to \pdty
+
+
*m
-
can
is
an
*
consecutive
cusps
"
the whole
lengthof
the
loopof
3ay2=x(x-a)2i.
the
curve
[OXFORD, 1889.]
a? that the lengthof the arc of the hyperbola xy and x=c is equal to the arc of the between the limits x=b the limits aV2 between curve r=b, r=c. ""2(a4+r4) 13. Show
=
=
[OXFORD, 1888.] 14. Show
that
in the
parabola
"=1+0080,-^ =:-__^_and T
hence
show
that the
extremity of
arc
'
between intercepted
the latus rectum
d"Y
sin^w*
and the_yertex is a{\/2-flog(l +\/2)}.
the
[I.C. S., 1882.]
ETC.
RECTIFICATION, 121.
Length
of the
Arc
of
an
143
E volute.
It has been shown (Diff.Gale, for Beg.,Art. that the difference between the radii of curvature
157) at
Fig. 12.
pointsof a arc corresponding
two
curve
is
equal to
lengthof
the
the
of the evolute ;
if ah be the arc of the evolute of i.e., of the original then (Fig.12) curve, .e.
(atA)
the
"
/"
portionAH (at H),
INTEGRAL
144
and
if the
CALCULUS.
volute be
regardedas a rigidcurve, and a from it,beingkept tight, then the stringbe unwound pointsof the unwinding stringdescribe a system of of which is the original AH. curves one curve parallel e
Find
Ex.
the
lengthof
the evolute
of the
Let a, a',/3, be the centres of curvature /3' the extremities of the axes, viz., A, A', B, B' of the evolute arc a/3 correspondsto the arc and we have (Fig.13) arc
[forrad. Thus
of
the
curv.
of
lengthof
a/2 /o(at 5)-p(at A) =
ellipse =
^.
the entire
=
~-
ellipse. correspondingto respectively.The of the curve,
AB
"
Ex. 3, p. 153,Diff. Beg."]. Calc.for
perimeter of
the evolute
EXAMPLE. in the above manner for the parabolay2 kax that the within the parabola lengthof the part of the evolute intercepted Show
=
is4a(3\/3-l). 122. Intrinsic The
Equation.
relation between
given curve,
measured
s,
the
from
lengthof the arc of a a given fixed point on
Fig. 14.
the curve, and the extremities of the of the
curve.
anglebetween arc
tangentsat the Intrinsic Equation
the
is called the
ETC.
RECTIFICATION, 123. To
the Intrinsic
obtain
145
Equation
from
the
Cartesian. be given as y=f(x). of the curve Let the equation and Supposethe #-axis to be a tangent at the origin, from the origin. the lengthof the arc to be measured
-^=/("),
(1)
s=\ *Jl+ [f'(x)~]2dx
(2)
Then
tan
also
from (2),and x If s be determined by integration eliminated between this result and equation (1),the requiredrelation between s and ^ will be obtained. .
Ex.
1.
be i/r
If
tangent
and
Intrinsic
the at the
equationof
the initial tangent at A and the have the radius of the circle, we
angle between point P,
and
circle.
a
a
therefore
Ex.
s"a^r.
the case equationis s
2.
=
For
of the
In
c
tan
tan^
catenary y
+
^ dx
and
/-,
as
T
"
dx P. I. c,
=
=
ccosh-, c
^. =
c
\/ 1 "
-r
"
i
""x
smh2-
=
c
K
the
trinsic in-
therefore
and
CALCUL
INTEGRAL
146
s
=
c
sinh
US.
-,
c
of the constant whence together,
being chosen integration s
124.
To
obtain
=
c
tan
so
that
x
and
s
vanish
\js,
the Intrinsic
Equation
from
the
Polar.
Fig. 16. /
",
to the the initial line parallel pointfrom which the arc is measured.
Take
usual notation
we
tangent at Then
the
with the
have
=/(0),the equationto 0+0, ^ T
the curve,
=
.......................................
.......
(1) (2)
from (4),and 0, "f" by integration of equations(2) and eliminated (3),the by means requiredrelation between s and \fswill be found. If
Ex.
Here
s
be
Find
found
equationof the r=a(l -cos 0).
the intrinsic
i/r
and a
sin
0
2
cardioide
CALCULUS,
INTEGRAL
148
the
125. When
By
dy
tan
have
we
of
means
of the Curve
Equation
^
=
-^ ax
is
given
d)'(t)
^_-^
=
.......
as
,-
.....
N
(1)
........
j (t)
equation(2)s
be
may
of t. the result and If then, between shall obtain the eliminated, we
found
by
tegrati in-
in terms
and
between
s
Ex.
In the
^.
cycloid y"a(\
we
have
tan
Also
whence Hence
126.
5=4a
5
=
equation(1) t be required relation
sin 4ot sin
t\
-cos
*
*mt
^
=
^ dt
=
tan
= ,
1+costf
2
+ cos02+ ax/(l
=
2a
cos
-,
2
if s he measured
-
sin2*
from
the
originwhere
Z=0.
is the equationrequired. T/T
Intrinsic
Equation
of the Evolute.
be the equationof the given curve. s=f(\f/) Let s' be the lengthof the arc of the evolute measured fixed pointA to any other pointQ. Let from some 0 and P be the pointson the original sponding correcurve to the pointsA, Q on the evolute; p0, p the P: at 0 and radii of curvature \j/the angle the tangent QP makes with OA produced,and ^ the anglethe tangentPT makes with the tangent at 0. Let
ETC.
RECTIFICATION, Then
149
and "*//-^r, =
ds
or
O
*
T
Fig. 18.
Equation of an Involute. if the curve the same With AQ be givenby figure, have the equation we s'=f(\}/), 127. Intrinsic
and whence Ex.
=
intrinsic
The
\Is "
\//,
\ equation
the
of
catenary
is s=ctsm\Ir
(Art. 123). Hence
and
radius
p0= =
' .
.
The
the evolute
c
p
its evolute
=
=
y
c
s
=
so
c
=
s
c
=
T/r=0
tan2^.
is
+ A^r + logsec T/T
that 5=0
measured s
and sec2i/r
I(ctan "fy + A)d^r
=
=
if s be
-
is
at the vertex
of curvature
1 ),or c(sec2-v//intrinsic equation of an involute
is
s
and
equationof
the intrinsic
c
T log(sec
when
constant
^=0,
; we
have
,
INTEGRAL
150
128.
Length of Arc
If p be the
CALCULUS.
of Pedal
Curve.
from perpendicular
originupon the tangentto any curve, and ^ the angle it makes with the initial line, we regardp, % as the current may polarcoordinates of a pointon the pedalcurve. culated Hence the lengthof the pedal curve be calmay by the formula
Ex. of the
Apply the above method to find the lengthof pedalof a circle with regard to a point on the a cardioide). (i.e.
Fig.
Here, if 2a
be the
arc
of
=
pedal
OP
=
the
-
J
/2a
cos
*dx
cos
-a
2
2
*
figure
2ctcos2*.
/2 A/a2cos4+ a2 sin2-
=
=
from
have
* cos
any arc circumference
19.
diameter, we p
Hence
the
=
4a sin
2 + C.
j
The
limits for the upper half of the Hence the whole perimeterof the
curve
1 2[4asin-Jo L-
2
are
pedal =8a.
x
=
0 and
X
=
TT.
ETC.
RECTIFICATION,
151
EXAMPLES. 1-. Find
the
length of
2. Find
the
lengthof the
of the
arc
any
fu\a x)=aP.
curve
"
[a,1888.]
completecycloidgivenby
y the
s
=
a
^
tan
equation of
^r+
sec
the
parabolais
+ sec ^). log(tani/r
a
Interpretthe expressions
the wherein closed given
Jw
0. 1
curve
that the intrinsic
4. Show
5.
cos
"a
for which the lengthof the arc measured the originvaries as the square root of the ordinate.
3. Find
from
a
=
6. The
line
taken
are integrals
rou
id the
a
[ST.JOHN'S, 1890.]
curve.
major
perimeterof
of
axis
is 1/10. ^/^eccentricity
is ellipse
an
1 foot
its circumference
Prove
nearly.
in
length,and
to be 3*1337
its feet
[TRINITY,1883.]
7. Show cardioide
that r
4r=3asec
0 remote
8. Find
length of the arc of that part of the cos 0), which lies on the side of the line from the pole,is equal to 4a. [OXFORD, 1888.]
the
a(l +
=
the
lengthof
of the cissoid
arc
an
r_asin26" cos
9. Find
the
10. Show
lengthof
show
a
that
of the
arc
certain
=
curve
equation of 4a(sec3Vr1).
that the intrinsic
3a3/2=2^ is 9s 11. In
any
ff
-
curve
5=ee\/2+
a
the semicubical
bola para-
Show
12.
is
CALCULUS.
INTEGRAL
152
the
that
given by
s
Show
13.
is 14.
s
Show
between
curve
y
of the
arc
+
that
the
the
curve
C.
Trace
tion equa-
a
of the
length
arc
of the
y=logcoth-
curve
points (xl9yj),(#2,3/2)is log s!n x^. sinn
15.
intrinsic
the
alogsec-
=
gd~l\ff.
a
=
an
=/("9)+/"(#)
in the
that
of
length
the
y2
curve
(a
=
g"
"
#)2,and
X-^
find
the
length
of that
od/
which
part of the evolute
the
corresponds to
loop.
[ST. JOHN'S, Find
the
of
of
and
1891.]
equiangular spiral pole. Show that the arcs of an from equiangular spiral measured the pole to the different with another points of its intersection pole but a different angle equiangular spiralhaving the same will form in series a [TRINITY, 1884.] geometrical progression. 16.
(p
17. has
measured
rsma)
=
length
1881
that
Show
from
the
for its intrinsic
arc
an
an
the
whose
curve
equation
s
is
pedal equation
p2=r2
a?
"
a"-.
=
Zi
18.
is
Show
equal
to the 19. curve
to
that that
the of
that
minimum
the
rm=amsinmO
length
ellipsewhose
an
and
maximum Prove
whole
of
semi-axes
radii of
length
the
the
limagon r=acos are equal
vectores
nth
of the
pedal
of
a
in
length
limacon.
loop of
the
is mn-m+1
,-m
(smmO)
a(mn+I)
m
dO.
^
1883
j
o
20.
Show
that
the
length
of
a
loop
of the
curve
[ST. JOHN'S, 1881.]
X.
CHAPTER
ETC.
QUADRATURE, 129.
Areas.
Cartesians.
bounded process of findingthe area is termed quadrature. portionof a curve The
It has bounded ordinates
by [x
considered
a
=
and
x
the limit of the
as
rectangles;and
two
area
"f"(x)], pair of any
=
the
sum
the
any
axis
of
an
that
of x9 may be infinite number
the
expression
is
area
the
[y
b] and
=
1 ydx
In
line
curved
any
of inscribed for the
in Art. 2 that
already shown
been
by
same
way
the
given abscissae [y
0 (x)dx.
or
=
bounded
area
c, y
d]
=
and
by any curve, the y-axisis
fxdy. by two Again,if the area desired be bounded given curves [y "p(%)and 2/ \^(^)]and two given ordinates \x a and x 6],it will be clear by similar reasoningthat this area may be also considered as the limit of the sum of a series of rectangles constructed 130.
=
=
"
=
INTEGRAL
154
indicated
as
in the
expressionfor the
figure. The be accordingly
will
area
CALCULUS.
Li% PQ
dx
fj"(0) \fs(x)]dx.
or
-
J
x=a
Fig. 20. Ex.
Find
1.
ordinates
the x"d
x=c,
Here
by the ellipse"- + 2
bounded
area
and
^2 =
-
1, the
b2
the
f ^Sr
area=
J
a
2a
For the
we quadrant of the ellipse above expressionbecomes
"
?
a2 .
givingirab Ex.
Find
2.
y2
area
the
area
x2 and
and
between
the
4
of the whole above
c=0
ellipse.
the #-axis
included
y2 ax. parabolatouch at the originand from #=0 are (a,a). So the limits of integration The area sought is therefore
curves
The
at
for the
and
.
2
2a
=a
^"
or
.
put d
must
a
=
%ax
"
circle and
=
the
fa
?
-
x2
~
cut to
x"a.
again
INTEGRAL
156
portionbetween
the
For
limits are For the
CALCULUS.
a
"
to
loopwe
the
0, and double
and
curve
the
asymptote the
before.
as
have
therefore
a+x
for the
portionbetween
the
and
curve
x\l /O
the
In
asymptote,
/v. _
dx.
" a+x
Fig.22. To
integrateIxJa~xdx, put J
*
cos
x=a
Then
a ~p x
and
0
dx"
"a
ftfJEfcfcl\ Va+x -
J
J
r =
and
sin 0 c?$.
area
of
a2/
--}" I)
1-cos2^
ETC.
QUADRATURE,
Again,
rxJ?E*dx=
sign before
of the
the
cos
"^gain fl^1 l-cos2#
negativesignis
radical
in
Bd9
"
J
"a+#
[The meaning +
a
-
J_
157
this
y=#/v/^" _
*
:
"
In
choosingthe
are
tracing the
we
a+x
the curve below the #-axis on the left of the origin and above the axis on the rightof the origin. Hence y being be is it referred to between limits the expected to, negative that we should obtain a negativevalue for the expression
portionof
Thus
the whole
area
requiredis
in this example that the greatest also be observed assumed that infinite one. In Art. 2 it was ordinate is an for the the result area finite. Is then was every ordinate ? t rue and the the bounded curve asymptote rigorously by limits let us integratebetween To examine this more closely small positive e is some a + e and quantity,so as 0, where have to exclude the infinite ordinate at the point x" "a, we
[It must
"
as
before
A/fEfdfc. [""c "
J
*a+x
where so
that 8 is
a
small angle. positive
This
4
which
close approaches indefinitely
when
8 is made
to diminish
without
is integral
2
to the former
limit.]
4
result
INTEGRAL
158
CALCULUS.
EXAMPLES. 1. Obtain
the
bounded
area
by
a
parabola and
its latus
rectum. 2. Obtain
bounded the areas by the curve, ordinates in the following cases : specified
the
and ^7-axis,
the
"
(a) #=ccosh-,
to x=h. x=
a
x=a
the
3. Obtain 4. Find
the
area
bounded
of X2la2+y2/b2l is divided areas
=
5. Find
the whole
area
the
to x=b. to x"b.
by the curves y2=4ax, #2=4ay. portionsinto which the ellipse
the line y=c. included between
by
the
curve
X2y2=a2(y2x2) "
and
its
asymptotes.
6. Find
the
between
area
the
curve
y2(a+x)=(a
"
xf
and
its
asymptote. 7. Find
131.
the
area
of the
Sectorial
Areas.
the
curve
y*x+ (x + af(x + 2a)
=
0.
Folars.
to be found
is bounded by a curve r=f(6) and two radii vectores drawn from the origin divide the area into elementary in givendirections, we small angle89, as shown in the sectors with the same figure.Let the area to be found be bounded by the arc When
the
loopof
area
PQ and the radii vectores OP, OQ. Draw radii vectores OP19OP2, OPn-i at equal angularintervals. Then by drawingwith centre 0 the successive circular arcs that the seen PN, P1NV P2^2,etc.,it may be at once of the circular sectors OPN, OP^N^ limit of the sum OP2N%, etc.,is the area required. For the remaining elements PNPV P^N^P^ P2^2P3,etc.,may be made to rotate about 0 so as to occupy new positionson the ...
ETC.
QUADRATURE,
159
greatestsector say OPn-iQ as indicated in the figure. Their sum is plainlyless than this sector ; and in the limit when the angle of the sector is indefinitely diminished
its
area
the radius
also diminishes
vector
OQ
without
remains
limit provided
finite.
O
Fig. 23.
The
of
area
a
circular sector is
of angleof sector. X circular meas. J(radius)2 the summation Thus the area required l?L"Zr2S(), being conducted for such values of 9 as lie between xOP and 0 6 xOPn-i, i.e., xOQ in the limit,Ox being =
=
=
the initial line. In and
the
xOQ
=
notation
of the
/3,this
will be
calculus integral expressedas
if xOP
=
a,
or
Ex. and
1.
Obtain
the
the initial line. the radius vector
Here
0=0
area
to
0"
-.
Hence
of the semicircle
sweeps
the
area
over
the
bounded
by
r
angularinterval
is
2
=
i.e., ^radius)*.
acos
from
0
CALCULUS.
INTEGRAL
160 Ex. 2.
Obtain
the
of
area
loopof
a
the
curve
sin 3ft
r"a
This curve will be found to consist of three equal loopsas indicated in the figure(Fig.24). The proper limits for making the integration extend over the first loop are 0=0 and 6 of 0 for which r vanishes.
of
.-. area
for these
=
-,
are
successive values
two
^
loop 1 fWn2
30 dO
=
=
f\l
60)d9
-cos
~
3~~ 12'
4
2
The
total
of the three
area
loopsis
therefore!^.
Fig. 24. EXAMPLES. the
Find 1. r2
=
bounded
areas
loopof 4. One loopof loopof r=asin2ft by the portionof r=ae^coiabounded 0=/3 + y (y being less than 2?r).
5. The
9=13 and 6.
Any
sector
of
7.
Any
sector
of
8.
Any
sector
of
cardioide
9. The
s
originand show
3. One
"2cos20+ 62sin2ft
2. One
10. If
by
that
be the
$
=
r=
a
sin 4ft
r
a
shift ft
=
radii vectores
7^0=^ ((9=a to 6*=^). r0""a (0=a to ^=)8). r(9
(9=a to a(l cos 0). a
=
r
=
0
=
fi).
"
lengthof =
between
r="tanh-
curve
the
2
the
27r,and A A
the
area
between
a(s air). "
the
same
points,
[OXFOKD,
1888.
]
ETC.
QUADRATURE, 132. Area
a
on
Closed
a
Curve.
(x,y) be the Cartesian coordinates of any point closed curve ; (x+ Sx,y + Sy)those of an adjacent
Let P
of
161
pointQ.
Let
(r,9),(r+
Sr , 6 +
$0)be the corresponding
shall suppose that in Also we polar coordinates. from P to Q along the along the curve travelling infinitesimal arc PQ the direction of rotation of the OP that the is counter-clockwise radius vector (i.e.
Fig.25.
hand to a person the left Then the element direction). is
area
on
AOPQ
ir2(S$
=
Hence
another is
curve
in this travelling
$(xSy ySx).
=
"
for expression
the
area
of
closed
a
f
Wxdy-ydx), beingsuch completelyround the the limits
133.
may
If
write
we
put
the above
y
that the
point(x,y) travels
once
curve.
=
i
so
that
^M^ =
(fo)we
where as ^\xzdv, expression
is
x
the limits of integrati that the current point(x,y) travels so chosen As v is really once completelyround the curve. 6 is a rightanglecare tan 6 and becomes infinitewhen be taken not to integrate must throughthe value oo to be
expressedin
terms
of
v
and
.
E. i. c.
L
INTEGRAL
162 Find
Ex.
by
CALCULUS.
this method
the
of the
area
ellipse
#2/a2+.y2/"2=l. Putting y
=
vx,
have
we
and
*
=
f"^L= f-
JV between
Now,
and
properlychosen
limits. in the first quadrant v varies from area
of
quadrant =?"
area
of
ellipse
therefore
134. If the
originlie
point P
current
elements such
L2
*
as
"
0 to
oo
Hence .
-,
=irab.
without
travels round
the
curve,
as
the
obtain
triangular of space such as OP1QV includingportions OP2Q2 shown in the figurewhich lie outside we
Fig.26.
the
curve.
removed travels element and
S6
These portionsare however ultimately from the whole integralwhen the point P the element over P2Q2, for the triangular
as OP2Q2 is reckoned negatively is negative.
135. If however
^ I(xdy
"
the
ydx), taken
longerrepresentsthe
curve
cross
round sum
the expression itself,
the whole
of the
9 is decreasing
areas
perimeter,no of the several
CALCULUS.
INTEGRAL
164
the curve is when speciallyadapted to the cases defined by other systems of coordinates. Ss of a plane curve, and OF If PQ be an element the chord the pole on from the perpendicular PQ,
Fig. 28.
|OF.PQ,and any sectorial area summation the along the whole being conducted of the IntegralCalculus In the notation bounding arc. AOPQ
=
=
this is
be at
[Thismay
deduced
once
from
| rW,
(V2d0ir^ds sin 0 \r =
=
(where "f"is
angle between
the
radius vector)
137.
Tangential-PolarForm. .
.
ds
Again, we
have
since
area
P
=
=
=
5^
d*p
P +
\ \pds
=
J
the
thus
:"
ds
tangent
and
the
ETC.
QUADRATURE, a
formula
suitable for
when
use
165
Tangential-Polar
the
equationis given. 138. Closed When
the
Curve. is closed this
curve
admits expression
of
simplification.
some
For and term
perimeterthe
the whole
in
round integrating disappears.Hence
when
the
curve
is dosed
first we
have area
^
=
the equationof Ex. C ale.forBeginners, By Ex. 23, p. 113, Diff. the one-cusped be the (i.e., cardioide) expressedas epicycloid may p
=
Fig. 29. Hence
its whole
limits
i/r 0 =
a2cos2^\d^ taken area=-^/f 9a2sin2;' "
and
^="
Putting-^ 3$,this =
and
doubled.
becomes IT
=
3a2
f (9sin2^ ^o
-
co**0)dO =
67ra2.
tween be-
INTEGRAL
166
Pedal
139.
Again,
CALCULUS.
Equation.
for
pedal equations,
have
we
A
ip
=
Ex.
ds
In the
Hence
dr
i p
=
sectorial
any
f
of curvature case
J
we
and
area
Ex. to the
1.
t
=
The
circle is
area
i
-
a,
rcosa
between evolute.
take
as
a
curve,
element
our
by
two
the infinitesimal
arc
infinitesimals
first order
sin
=
y2si
elementary trianglecontained of curvature
0 dr
sec
/
included and the
140. Area
this
}p
area
/"2
In
=
equiangularspiralp=r
=
To
their
given by
curves
.e.
p, between
(Fig.31)
a
this
p\
is
radii
two
of
the
area
contiguousradii ds of the
curve.
and |/o2"S\^,
the
or
its involute,and circle,
a
tangent
ETC.
QUADRATURE,
167
the tractrix and its asymptote is between Ex. 2. The area found in a similar manner. such that the portionof its tangent The tractrix is a curve and the ^7-axis is of constant between the point of contact
length c.
Fig.31.
Taking two adjacenttangents and elemental triangle (Fig.32)
o
the axis of
T
x
as
forming an
r
Fig. 32.
EXAMPLES. 1. Find
the
area
of the
[Limits\jf 0 =
2. Obtain
the
same
two-cuspedepicycloid
"^=7rfor
to
result
by
means
7.2 ^2 + =
[Limitsr=a
to
r
=
one
quadrant.]
of its
pedalequation
1^2.
2a for
one
quadrant.]
CALCULUS.
INTEGRAL
168 3. Find
the
the radius
between
area
of curvature
the
at
catenary s the vertex, and
=
c
^,
tan
any
its
other
evolute, of
radius
curvature.
the
4. Find
evolute,and 5. Find
evolute,and
area
any
between
the
epicycloids
=
AsmB^s,
its
equiangularspirals"Ae^y
its
radii of curvature.
two
AREAS of Pedal
141. Area
the
radii of curvature.
two
any
the
between
area
OF
PEDALS.
Curve.
If
_p=/(Vr)be the tangential-polar equation (Diff. Gale, for Beginners,Art. 130) of a given curve, S\fs will be the angle between the perpendiculars two on contiguoustangents,and the area of the pedal may be expressedas
(compare Art. J|p2c^/r
131).
Fig.33. Ex.
Find
the
area
of the
pedal of
a
circle with
regardto
a
(thecardioide). pointon the circumference if OF be the perpendicular Here the tangent at P, and on OA the diameter obvious that OP ( 2a), it is geometrically =
bisects the the
Hence, callingYOA"^, angle AOY. tangential polarequation of the circle
Hence
=
^/
we
have
for
ETC.
QUADRATURE,
169
where the limits are to be taken as 0 and TT, and the result to be doubled so as to include the lower portionof the pedal. Thus
*cos*fe^ 4a2. ^l=4aaf 2 J =
o
2
f 4222
J o
Fig. 34.
142. Locus Let
0 be
a
o
Origins of
Pedals
of
given
Area.
fixed
ordinates point. Let pt \jsbe the polarcothe foot of a perpendicular OF upon any givencurve.
a
of
tangentto
of
0
Fig.35.
Let P
be any from
other P
upon
fixed the
point,J"F1(=^1)the perpendicul tangent. Then the areas
INTEGRAL
170 of the
CALCULUS.
0 and
pedalswith
P
as originsare respectively
and j[ftL2"% ijV2cfyr taken
between A
and Al coordinates of P areas
Cartesian Pi and p is 2 Al
Call limits. respectivelyLet r, 6 be the with
regard
equivalents.Then T cos($" -t/r) P =p ~~
"
a
function
known
\(p \p^d\fr
=
definite
certain
=
"
x
of
cos
x
"
cos
ifs "
x9
y sin
y
their
i/r,
\fs Hence
i/r "
"
y sin
\l/fd\^
"
1cos + x* Icos2i/r + 2a32/ c?i/r
si \[s
1si 2/-2
+
2
0, and
polar
Vp^d"^2x \pcos ^ d\/r 2^/|psin \fsd\fs
=
Now
to
these
I 2 Ipsin \/r d\[s, d\fs, \/r
Jp
cos
such limits that the whole pedal is described Call them will be definite constants.
between
-20, and
2Al
=
If then P move its locus must
"2A +
Hence
a
known
2gx + 2fy+
2hxy + by2. that Al is constant,
ax2 +
in such a manner be a conic section.
143. Character It is
2A, 6,
a,
obtain
thus
we
-2/,
of Conic, result in
it will be obvious
that inequalities
that if p, q, r,
...
stand ,
for
INTEGRAL
172
will
removed.
thereby be
if II be the
vanish, and
for any
have
we
2A1 generalcase.
in the
"2 is
Thus
of the
area
point such
211 + ax2 +
=
The
of this conic is
area
"
*/ab
,
""
"
s
"
I?
\
(area or conic).
, --
Thus
h2
"
TT
-d.1 IH =
the
pole
2hxy + by*
(Smith'sConic Sections,Art. 171). A
pedal whose
other
*
For
a
both and \psiu\^d\[^ integrals\pcos\fsd\fs
that the
is "2
CALCULUS,
^7T
of any
case particular
closed oval the equation
of the conic becomes
whence
J.1
where
is the radius of the circle
r
values of
constant
on
which
i.e.the distance of P
Av
P
from
146. Position
of the
Point
"2 for Centric
In
which
has
centre
oval
any
plainlyat
a
that centre, for when
the
the centre
lies for "2.
Oval.
point "2 is taken
is as
and \psui\fsd\ origin,the integrals\pcos\fsd\fs both
vanish
complete oval cancelling), 147-
Ex.
Here and Hence
1.
Find
the
the
area
point within (a limagon).
regard to centre
is performedfor the integration (oppositeelements of the integration
when
the
any
A n
=
=
of the pedal of a circle with circle at a distance c from the
n+^, 7ra2.
Ai=ica*+"
.
ETC.
QUADRATURE,
of the pedal of an to any point at distance c from the centre. of the pedalwith In this case II is the area Ex.
Find
2.
the
/* Vcos2"9 + b%m*0)dO
2
-
area
Ex.
c
with regard ellipse
regardto
the centre
"2)|.
(a2+
^1=|(
Hence
taken
=
173
The
3.
with
from
area
pedalof the cardioide r=a(l "cos 0) internal pointon the axis at a distance
of the
respectto
an
the poleis
|(5"s-2"c+2c'). [MATH.TBIpos"187a] pole,P the given internal point; p OF2 and PTl on any tangent perpendiculars
Let 0 the two
and P pl "p
be
respectively ; ""the angle Y$P "
c cos
and pl from 0
the
"",and
^Al=2A
"
and
2clp
cos
OP"c
; then
+ / ""cfc"
Fig. 36. Now
in order
that p may
between integrate the cardioide
limits
sweep
"" =
0 and
the whole
out
"" =
-^and
pedalwe
double.
Now
must
in
(Fig.36) p=
OQ
sin
Y2QO
=
OQ
sin^xOQ. [Dif.Calc.,p. 190.]
CALCULUS.
INTEGRAL
174
For
r2"0
itf0"| =
=
|-{*-("-W-|
Hence
|-*=f,
or
J-J-J,
-
/3
so
"
-
-.
23
A*-
,
Hence
/p
cos
"j" d"j" 2 =
,
2a cos3 2
/
cos
d(f" "/"
cos
3,so?2
'
=
fl
4a
3
x
/ cos% o
=
rf [4cos%
/
12#
3
-
cos")"iz
42
6422
Also
2
^^^ fc2cos2d"cta=3.2c2i J
222 Sir
Finally
24
=
2
Tcos^"fe, 4a*"*^*J"
J
=
6
1
3
642 mi
A Al
When
f2 is taken 2A
=
l
Hence
as
it is positive,
'
T"
8~
of Minimum
Area.
it appears origin,
that
Pedal as
211 +
the term
?rac
--
Origin for
^
2
_?ra
148.
24a2
J(05
cos
^
+ y sin
\Jsfd\ls.
is necessarily \(xcos\fs + ysm\}s)"2d\fs
clear that
Al
can
never
be
less than
II.
ETC.
QUADRATURE, "2 is therefore
pedal curve 149.
has
Pedal
in Art. 138
minimum
a
of
which
corresponding
the
area.
Evolute
an
for the
formula
The
originfor
the
175
of
Closed
a
of any
area
Oval.
proved
closed oval
is of
area
jp2*/' Jjft)
Hence
J
=
oval +
plainlyexpresses that the area of any pedal of the oval itself is equal to the area of an oval curve togetherwith the area of the pedalof the evolute (for which
-ry
is the
radius
also admits
This
vector
of
area
of
above
article shows
pedalof
evolute
the
evolute).
proof. elementarygeometrical
of the Find the area Ex. with regardto the centre. The
pedalof
of the
=
-
pedalof
the
evolute
of
an
ellipse
that area
of
pedal of ellipse area -
-(a2+ b2) -
irab
=
?(a b)2. -
of
ellipse
INTEGRAL
176 150. Area of
pair
bounded
by
a
Curve, its Pedal, and
a
Tangents.
P, Q be
Let
CALCULUS.
two
contiguouspoints on a given 7, F' the corresponding pointsof the pedal of curve, since (with the usual notation) Then any origin0.
PF=-vjrthe elementarytrianglebounded contiguoustangentsPY, QY'
and
the chord
by YY'
two
is to
the firstorder of infinitesimals
Fig.38.
Hence
the
area
curves
and
a
may
be
and
is the of the
area
151. Let
of any portionbounded by the two curve pair of tangentsto the original
expressedas
same
as
the
pedal of the
portion of the corresponding evolute.
Corresponding Points f(x,2/) =
0 be
any
and
closed
Areas.
curve.
Its
area
(A^)
QUADRATURE, is
177
\ydxtaken line-integral
the
expressedby
ETC.
round
the complete contour. If the coordinates connected relations the curve
by
of the current point (x,y) be with those of a second point(" rj) by the mg, y nrj, this second pointwill trace out is expressed /(w" nrj) 0 whose area (J.9)
x
=
=
=
Irjdg taken line-integral
the
And
its contour.
have
we
\ydx
l=
whence
round
=
\nrjm
=
it appears that the area of any 0 is mn times that of f(inx, ny)
f(xty)
=
152. Ex.
1.
Apply
this method
**
,"*
+
~
=, a
b
r
to find the
=
curve
0.
of the
area
ellipse
1
r
the
point", T\ traces corresponding
and
area
of
closed
out the circle
ellipse
x area
=~
of circle
r
Ex.
Find
2.
the
area
Let
in
curve
^
ny
mx
then the
or
of the =
the central E. i, c.
pedal of
r2
=
n^f)2 =
a
V
+
ij,
^- cos2 0 + m2
an
=
+
is
correspondingcurve
polars
(mV
sin20, ri2 "--,
symmetrical about ellipse, M
both
axes.
Wif-
INTEGRAL
178 Hence
the
CALCULUS.
of the first curve
area
=
x
"
of second
area
mn
EXAMPLES, 1. Find
the
of the
area
loopof
the
curve
ay"L=x\a-x). 2. Find
the whole
of the
area
[I.C. S.,1882.]
curve
a?y2 a2x2-x*. y\ and a2#2=;?/3(2ct
[I.C. S., 1881.]
=
3. Trace
the
curve
that of the circle whose
equalto
that its
prove
"
radius is
the curve cfiy*xb(Za x\ and prove radius is a as 5 to 4. to that of the circle whose 5. Find
=
the whole
and
that its
"
of the
area
is
a.
[I.C. S., 1887 4. Trace
area
1890.] area
is
curve
[CLARE, etc.,1892.] 6.
By
of the
show
means
integral/y
of the
formed triangle
that
by
they enclose
the
the
taken
dx
round
the
contour
lines intersecting
area
[Sir.PKIZE, 1876.] 7. Find
the
area
between
and
y2
=
a
8. If
show
ty be
that the
the area
of y
an
oval
cos
ds "fy
integration being taken
makes
with
the axis of #,
is
curve
or
its asymptote.
x
angle the tangent
/r the
"
q:
all round
/x the
sin
^rds,
perimeter.
INTEGRAL
180
CALCULUS.
line to a point P on the curve if A be the ; and by the curve, the initial line,and the radius vector 9,42 Find
21.
the
P,
to
then
2rf.
=
of the closed
area
bounded
area
3a sin
portionof 0
the Folium
9
cos
_
~sin^6"TcoW ratio does
what
In
the
line x+y
[I.C. S., 1881] divide
Za
=
the
loop? 22.
of the
area
[OXFORD, 1889.] Find
the
of the
area
given radii vectores
and
r=aOebe
curve
between
enclosed
branches
successive
two
of the
two
curve.
[TRINITY,1881.] 23.
Find
the
24. Show
^" -,
and
that
the
of
area
the total
state
loop of
of the
area
the
loop of
a
in the
area
curve
r
a0cos 0 between
=
the
curve
n
odd, n
cases
r
=
acosn0
is
even.
4?i
25.
Find
the
of
area
loop of
a
the
curve
r
=
a
cos
3$ + b sin 3$.
[I.C. S., 1890.] that the 26. Show the curve r=acos5$
contained
area
is
between
the circle
equal to three-fourths
of the
circle. that the
of the
area
-
equal to
curve
30.
its
the =
Find r
=
9+
cos
a2sin2#) aV =
[I.C. S., 1879.]
irac.
Find
equation r 29.
area
curve
sin 0
2ac r2(2c2cos2"9
28.
of the
[OXFORD, 1888.]
27. Prove
is
and
r=a
whole
0+
acos
the
of
area
the
b,assuming included
area
b
"
a.
between
the
two
a(2cos 0 + ^3).
Find
the
loops of [OXFORD,
between
area
the
the
representedby
curve
1889.
r=a(sec $+cos 0)
curve
the
]
and
asymptote.
31.
Prove
lemniscate
that r2
=
the
a2cos2$
area
of
with
respect to the
loop
one
of
the
pedal
of
the
poleis a2. [OXFORD, 1885.]
32.
Find
the
area
of the
loop of
the
curve
(x'\-y)(x^+y2)^axy.
[OXFORD,
=
33.
Prove
that the
area
of the
loop of
the
curve
1890.]
QUADRATURE, Find
34.
and
its
35.
the whole
that
between
vector
r
36.
between
the
of the
Show
the
the
the
centre, is
a2+b2-r2 ellipse(*L^ p2
the
semi-major axis,and
tan"1^/^^-, a,
"
0 and
=
its
between
being
the
the
a2log(sec ""+
-
curve
5
=
atan^,
V*"""""ig
at
tangent
^ + a2 tan "/"
-a2 tan
b
radius
a
[CLARE, etc.,1882.]
in eluded
area
in-
=
ellipse.
^
at
of
area
curve,
that the
tangent
curve
[OXFORD, 1888.]
the
from
semi-axes
its
contained
area
181
asymptotes.
Show
eluded
ETC.
tan
c"). [TRINITY, 1892.]
37. Show p =^isin 38.
that
Sty and
Show
whose
the
the epicycloid space between taken from cusp to cusp is ^irA2B.
its pedalcurve
that
areas
the
of
area
the
a2(fTT
are
a(^\/3 + cos^#) has three loops 2\/3), a2(f f\/3), spectively. a\ -far fV") re-
curve
r
+
=
TT
-
-
[COLLEGES, 1892.] 39. Find
the
of
area
loop of the curve x*+y* Za2xy.
a
[OXFORD, 1888.]
=
40.
Find
the
of the
area
the
pedal of
curve
d*)l, =*("**the
originbeing taken
41.
the
Find
curve
42.
the
3%2
Find
at
x
=
*Ja2 62,y "
included
area
between
and a2(#2+;?/2)
=
the whole
its
of the
area
=
0. one
[OXFORD, 1888.] of the
asymptotes.
Find
the
curve
tf+yi a\x*+y*). of a loop of the curve (mV + n,y)* aV b2y2-
area
=
44. areas
Trace
the
shape
of the
of
[a,1887.]
[a,1887.]
=
43.
branches
-
followingcurves,
[ST.JOHN'S, 1887. ] and
find
their
: "
(i.)(^+^2)3 =aay*.
(^2+ 2/)3 a^?/4. (ii.) =
[BELL, etc., SCHOLARSHIPS, 45.
Prove
that the 3?
V2 '
of
area '
1 / X2
V2\2
"
7TC2/
1887.]
INTEGRAL
182
46.
Prove
that
the
CALCULUS.
in
area
the
positive
(av+w^w
of
quadrant
the
curve
^(5+5).
is
[a;18900 47.
Prove
that
the
of
the
f")
is
area
curve
-3""
(V2
+
-
tan-1
a2)
-
.
[ST. 48.
Prove
that
the
of
area
the
JOHN'S,
1883.]
curve
9,aV h
62 where 49.
is
c
is
Prove
less
both
than
that
the
area
a
and of
5, the
is
7r(ab
"
Prove
[OXFORD,
^4-3o^3
curve
fTrtt2. 50.
c2). +
[MATH. that
the
areas
of
the
loops
two
24^3)
are
(32^
and
(167r-24\/3)a2,
+
of
the
1890.]
a2(2^2+y2)=0 TRIPOS,
1893.]
curve
a2,
[MATH.
TRIPOS,
1875.]
CHAPTER
SUKFACES
XI.
OF
SOLIDS
about
the
VOLUMES
AND
OF
KEVOLUTION.
It
of Revolution
Volumes
153.
revolve
about
ordinates
x
=
Art.
in
shown
was
the
of
axis
x^ and
x
that
5
the
x
is to
x2
=
if the
a"axis.
curve
y=f(x)
portionbetween be obtained by
the the
formula *2
dx.
Tr2 .
154. More
AB,
and
About
any
axis.
generally,if if PN
be
the
any
revolution
be about
perpendiculardrawn
any
line
from
a
CALCULUS.
INTEGRAL
184
point P on the curve upon the line the volume contiguous perpendicular, if 0 be
or
155. Ex.
loopof
the
Here
1.
Find
the volume
formed
volume
=/
J
J
Let the Then and
a
5a2z
-
+
the revolution
3) about
of the
the tf-axis.
+ x
2az2
~
3 _J"
tion of the spindleformed by the revoluabout the line joiningthe vertex to one
the volume
parabolicarc
extremity of
as expressed
this becomes
z
Find
a
o
rf2a3 log Ex. 2. of a
P'N'
I x*a~xdx. 7ry2dx=7r J
o
Puttinga +x=z,
by
f=x2?"^ (Art.130, Ex.
curve
is
and
the line AB
givenpointon
a
AB
the latus rectum.
parabolabe
the axis of revolution
Fig.
40.
y2
4o
=
is y
=
2^7,
P
"fi
REVOLUTION.
OF
VOLUME
185
Also
and
volume
dAN
= .
o
4?r
75
156.
Surfaces
Aain, if S be out
of Revolution. the curved
y the revolution
of any
surface of the solid traced arc
AB
about
the
^c-axis,
Fig. 41.
PN, QM two adjacentordinates,PN being the of the smaller,3s the elementaryarc PQ, SS the area
suppose
elementaryzone
traced
out
by
the revolution
of
PQ
CALCULUS.
INTEGRAL
186
about
the
#-axis, y and
ordinates of P and Q. Now take we may out
would
be if each
it
axiomatic
as
point of
it
the axis, and less than distance QM from the axis.
SS lies between
therefore
in the limit
8s and
^y
This may
as
from
greaterthan
it
distance PN
pointwere
at
a
2w(y + Sy)Ss,and
8
2-7T2/
"
\
as
to
be
values
of
in any
convenient
particular
beingobtained -j-" -^, etc.,
-r-"
the differential calculus.
157. Ex. formed
area
f
be written
example,the
the
the
have
we
or
=
happen
may
that
at the
if each
r/^ -j-
is
were
from
Then
lengths of
the
in its revolution
by PQ
traced
Sy
y +
by
Find
1.
the revolution
Here
surface
the
of the
of
a
belt
y2
curve
=
"ax
of the
paraboloid
about
the #-axis.
about
the
=
V x
dx
dx "/"
y"dx /X"1 dx
Ex. line. Here
The
2.
Find
curve
r
the volume
volume
=
=
=
a(l+
and
/try^dx =
TT
cos
surface TT
6}
revolves
of the
figureformed.
/?'2sin20d(rcos 0)
/a'2(l 0 -f cos2$), -{-cos #)2sin2#a c/(cos
initial
INTEGRAL
188 2. A
Show
and
CALCULUS.
of radius a, revolves round quadrant of a circle, that the surface of the spindlegenerated
that its volume
=
its chord.
-^-(10 3?r). -
3. The
part of the parabolayL "ax cut off by the latus revolves about the tangent at the vertex. Find the surface and the volume of the reel thus generated.
rectum
curved
=
THEOREMS 158. I. When line in its own
OF
any
PAPPUS closed
OR
curve
GULDIN. revolves
about
a
plane,which does not cut the curve, the volume of the ring formed is equal to that of whose height and a cylinderwhose base is the curve is the lengthof the path of the centroid of the area of the curve. Let the #-axis be the axis of rotation. area
with
Divide
the
elements up into infinitesimal rectangular sides parallel such as to the coordinate axes,
(A)
Fig. 43.
each PjPgPgP^,
of
area
SA.
Let the ordinate
PlNl y. infinitesimal angle
Let rotation take placethrough an 89. Then the elementarysolid formed
is
on
=
base
SA
its height to first order infinitesimals is ySO,and therefore to infinitesimals of the third order its volume and
is SA
.
THEOREMS
If the
OF
be
rotation
PAPPUS.
through
obtain by summation SA y If this be integratedover have for the volume curve we .
.
189
finite
any
angle a
we
a.
the
whole
area
of the
of the solid formed
a!i/cL4. Now of
a
for the ordinate the formula of masses number m2, ..., mv
of the centroid with
ordinates
X?7? II
2/i"2/2'""""
-^ then
is y=
seek
we
"
y
the value
of
of the curve, each the ordinate of centroid of the area element 8A is to be multipliedby its ordinate and the sum of all such productsformed, and divided by the
or
of the elements,and
sum
in the
language of
the
have
we
Calculus Integral
A (yd (yd A J
y
=
_
=
i -
.
\dA
A
Thus Therefore But is the
A
volume is the
formed of the area
lengthof the
=
path
A(ciy). revolvingfigureand
This establishes the theorem. COR. If the curve perform have and form a solid ring,we a
=
2-7T and
ay
of its centroid.
volume
complete revolution,
a
=
A(2jry).
closed curve revolves about a 159. II. When any line in its own plane which does not cut the curve, the curved surface of the ring formed is equalto that
INTEGRAL
190
CALCULUS.
and whose of the cylinderwhose base is the curve heightis the lengthof the path of the centroid of the perimeter of the curve. Let the #-axis be the axis of rotation. Divide the perimeters up into infinitesimal elements such as PXP2 each of lengthSs. Let the ordinate PlNl be called y. Let rotation take placethroughan infinitesimal angle S9. Then the elementary formed is ultimately area a with rectangle of the second
sides Ss and ySO,and to infinitesimals order its area is Ss y"9. .
Fig. 44.
If the rotation be through any finite angle a we obtain by summation Ss ya. If this be integrated the whole over perimeterof the curve have for the curved surface of the solid we formed .
an/cfe. If
of the seek the value of the ordinate (rj) centroid of the perimeterof the curve, each element of Ss is to be multiplied by its ordinate,and the sum we
THEOREMS
OF
divided
productsformed, and
all such
the elements,and
US.
PAPP
191
by
the
sum
of
have
we
Lt
languageof
in the
or
Calculus Integral
the
^yds\yds n
\ds
\yd8=8tj,
Thus the surface formed
and
s
s(afj).
=
and perimeterof the revolvingfigure, arj is the length of the path of the centroid of the perimeter. But
is the
s
This establishes the theorem. Con. If the curve perform form
and
solid
a
have
ring,we surface
completerevolution
a =
a
2?r and
s(2 -73-77).
=
and surface of an anchor-ringformed by radius about of circle a line in the plane of a a the circle at distance d from the centre are respectively Ex.
The
volume
the revolution
of
volume surface
Tra2X 2?rc? 27T2a2o?, =
=
2:ra
=
x
Zird
=
4ir2ad.
EXAMPLES. 1. An
major 2. A an
revolves ellipse
axis.
Find
extremity
volume 3. A
which
the volume
revolves
square
about
of the
about
other
the tangent at the of the surface formed.
end
of the
to a diagonalthrough parallel diagonal. Find the surface and a
formed. scalene does not
surface and
trianglerevolves about any line in its plane the cut triangle. Find expressionsfor the
volume
of the solid thus formed.
INTEGRAL
192
Revolution
160. When
we
OAB
area
the
Area.
|r"2o0.Being ultimatelya centroid is f of the way in a completerevolution
from
such
up into OPQ, whose
as
infinitesimals
by
triangularelement, its 0 along its median, and
the centroid
r sin 6) 27r(f
the
revolvingarea
to first order
denoted
about
revolves
sectorial elements
be
may
Sectorial
a
divide
may
infinitesimal area
of
sectorial
any
initial line
CALCULUS.
or
travels
a
distance
the volume
traced
f irr sin 9.
Fig. 45.
Thus
by
Guldin's
by
of this element
the revolution
to first order
traced
by
first theorem
and infinitesimals,
161. If x we
we
=
OAB
sin 9 d9. 7r[r3si
y
=
rsin9,
r3sin 9 S9
have
area
the volume
put
rcos9,
=
therefore
of the whole
the revolution
f
is
r3sin 9 .
7
and
=
r3sin
=
r*cos*9t St
"
=
(9$(tanlf) ~
=
xH
St,
is
EXAMPLES.
and the volume
193
therefore be
may
as expressed
(xHdt. EXAMPLES. and surface of the right the volume 1. Find by integration circular cone formed by the revolution of a right-angled triangle about a side which contains the rightangle. 2. Determine
generatedby
the entire volume of the ellipsoid which is the revolution of an ellipse around its axis major.
[I.C. S.,1887.]
that the volume of the solid generated by the round revolution of an its minor axis,is a mean ellipse portional probetween of the those generatedby the revolution and of the auxiliary circle round the major axis. ellipse 3. Prove
[I.C. S., 1881.] that the surface of the prolate 4. Prove spheroidformed by the revolution of an of its major e about ellipse eccentricity axis is equal to 2
.
of
area
ellipse .
formed Prove also that of all prolate spheroids of surface.
ellipseof
an
the revolution the greatest
by
sphere has
[I.C. S., 1891.]
5. Find
the
given
area, the
of the solid
the volume
loopof
the
y^"x^
curve
producedby
about
the revolution
the axis of
of
x.
[I.C. S., 1892.] 6. Find revolution
the surface and volume of the reel formed of the cycloid round a tangent at the vertex
7. Show that the volume of the cissoid y2(2a "
of the solid formed tion by the revolu^)=x3 about its asymptote is equal
to 2?r2a3. 8. Find
the
curve
the
by
[TRINITY,1886.] the volume of the solid formed by the revolution of (a x)y2 a?x about its asymptote. [I.C. S. 1883. ] -
=
,
9. If the
show
curve
r
=
that the volume greaterthan b. E. i. c.
a
+ b
cos
0 revolve
about
the
initial
line,
4- b2)provided a generated is "7ra(a2
be
[a,1884.] N
CALCULUS.
INTEGRAL
194
the
Find
10.
about
the
and
that
Show
11.
The
if the
of
loop
by
the
the
within
lying
area
the
curve
the
Zay2=x(x
curve
Find
y=a.
revolution r^
=
1890.]
[OxTOKD,
=
of
loop line
straight
formed
parabola r(l+cos $) 2a, volume generated is 187ra3.
line, the
12.
solid
the
the
without
initial
of
0=|.
and
0
=
the
of
radius
prime 6
between
volume
the
volume
"
of
cardioide
revolves
about
the
1892.]
[TRINITY,
1890.]
[OXFORD, Show
13.
of
area
that
the bounded
r=f(0)
of
coordinates the
by
the
the
about a)2 revolves the solid generated.
centroid
0=a,
vectors
the
sectorial
ft,has
for
of
6
=
its
coordinates
f 14.
on
the
Show
the
that
initial
line
at
centroid
a
of
the
distance
cardioide
from
the
-
r
a(l"
=
$)
cos
is
origin.
6 15.
cardioide
the
If
p=rcos(9
"
y\
prove
r
that
=
a(l the
3^7r%2 16.
about volume
cos
#)
volume +
revolve
round
generated
f 7T2"3cos
[ST. JOHN'S,
y.
curve
27r2a3(l+e2). 17.
The
lemniscate
r2
=
a2cos2#
revolves
pole.
Show
that
the
volume
generated
line
1882.]
revolves that
the
[I. C. S., 1892.1 about _
the
the
is
is very 0\ where small, r=a(l -ecos e the initial line. Prove to parallel tangent of the solid thus generated is approximately
The a
"
is -
a
tangent
at
INTEGRAL
196
CALCULUS.
of the
is Sx.Sy, and its mass rectangleRSTU be regarded (to the second order of smallness) may as 0(0,y)Sx Sy. Then the mass of the stripPNMV may be written area
in
or
conformitywith
the
notation
of the
Integral
Calculus
between
the limits y
y =f(x). In performing this integration (withregard to y) x is to be regarded
the
masses
i.e.the
mass
If then all such
we
we
search
are
for the
mass
may
AJKB
area
be summed which the above must the ordinates AJ, BK, and the result may
be written
with the limits of the integration b. from x a to x =
of the
stripsas
lie between be written
which
0 and
findingthe limit of the sum of of all elements in the elementarystripPM, of the strip PM.
constant, for
as
=
=
regard to
x
being
INTEGRATION.
DOUBLE
Thus
mass
of
197
area
A JKB
=
n
or
164. Notation. This is often written
ff "j"(x,y)dxdy, order. the elements dx, dy beingwritten in the reverse There is no uniformlyaccepted convention as to the order to be observed,but as the latter appears to be shall in the the more used notation,we frequently presentvolume adoptit and write
'x,y)dxdy when with
we
are
to be made to consider the firstintegration
regardto
y and the second with
when the firstintegration is with is to say, the right-hand element
regardto
regardto
x.
x,
and
That
indicates the first
integration. If the surface-density of a circular disc bounded by xP+y2 a2 be given to vary as the square of the distance from the y-axis, find the mass of the disc. "JL^ ^ Here we have [juv2 of the element 8x 6y,and its for the .mass is therefore /*#2"#6y, and the whole mass will be mass Ex.
=
// limits for y w411 be ;?/ 0 to y=*Jdl xL for the positive and for x from #=0 The result must then to x=a. quadrant,
The
=
"
INTEGRAL
198 be
multipliedby 4, for the four quadrantsthe the first quadrant.
CALCULUS.
the
distribution
being symmetrical in
of the whole
mass
is four times
that of
Fig.47. and
Putting x=asm0
165.
Other
The
same
dx=acosOdO,
Uses
of Double
theorem
may
be
we
have
Integrals.
for many other few illustrative examples, used
purposes, of which we givea which may the field to indicate to the student serve of investigation But our now scope in open to him. the presentwork does not admit exhaustive treatment of the subjects introduced.
DO
Find
Ex.
UBLE
TION.
INTEGRA
of
the statical moment r2
199
quadrant of
a
the
ellipse
4,2
_+"_ 1,9 a29 62
=
!
'
the surface-density beingsupposeduniform. y-axis, Here each element of area 8x8y is to be multipliedby its surface densitycr (which is by hypothesisconstant in the case and the sum supposed)and by its distance x from the y-axis, the whole of such elementary quantitiesis to be found over will from The be limits of the quadrant. integration y=0 to about
the
7
y
_Va2
=
"
x* for ?/ ; and
from
#=0
to
for
x=a
x.
Thus
we
have
a
moment
/
=
crxdxd'u=^"\Wa2
/
J J
a
"
x2dx
) 0
00
_"rbr _(a?-x2)%-\a_o-ba* aL
Gentroids.
166.
formulae
The
3
Cartesians.
proved
of the centroid of at
Jo
3
in statics for the coordinates
of
number
a
masses
mv
m2, m3,
.
.
.
,
points(xv y^, (x2,y2),etc.,are "
_ ~
apply these to find the coordinates of the centroid of a given area. (See also Arts. 158, 159.) be the surface-density For if at a given point, We
may
o-
then
or
Sx
Sy is the
of the
mass
-
_ "
or,
as
it may
be written
element,and
S(crox 8y)x I("rSxSy)9 when
the limit is taken
I dy \\" arx
dx
J\ardxdy
INTEGRAL
200
CALCULUS,
jja-ydxdy Similarly
~
ff~
\\a-dxdy j J
the limits of
integration beingdetermined will
summation
be
effected for
the
that the
so
whole
in
area
question. Find Art.
the
centroid
of the
quadrant elliptic
of the
Example
in
165.
It
proved
was
moments
there about
/ /"rdxdy=
Also Hence
that the
y
167. Moments When
=
limit
y-axiswas
of the
of the
sum
mentary ele-
?" "
.
quadrant=^^-.
of the
mass
*="
Similarly
the
=-"
3/4
STT
"
of Inertia.
is multiplied by the every element of mass of its distance from a given line,the limit of of of such products is called the Moment
square
the sum Inertia with regardto the line. Such quantities of greatimportancein are Ex.
Dynamics.
Find
the moment of inertia of the portionof the parabola f/2 4a# bounded by the axis and the latus rectum, about the #-axis supposing the surface-density at each pointto vary the nth power of the abscissa. as Here the element of mass is =
/x being a
constant,and the Lt
where
moment
2/i#V*"a? 8y
the limits for y
are
from
or
0 to
of inertia is //,\
\y*xndxdy,
2vW, and
for
x
from
0 to
a.
INTEGRATION.
DOUBLE
We
thus
201
get
Mom.
In.
"
=
=
3
" f* oj
o
U
3
Again,the
of this
mass
+
fo
portionof
ny\/ax
the
parabolais givenby Ca\~
~~l
l*xndxdy p\ \y\ =
xndx
--
271 + 3
Thus
have
we
In. about
Mom.
0ff=3
EXAMPLES. the the first quadrant of the circle ,272+^2=a2 densityvaries at each pointas xy. Find of the quadrant, (i.)the mass " its of gravity, centre (ii.) .("") its moment of inertia about the #-axis. (iii.) 1. In
2. Work
out
the
results corresponding
parabolay^=^ax bounded by the varying as xpyq. surface-density
for the
surface
portionof
the
the latus rectum, the
axis and
varies centroid of a rod of which the line-density the distance from one end. as Find also the moments of inertia of this rod about each end and about the middle point. 3. Find
the
4. Find
the
centroid of the trianglebounded by the lines at each the surface-density y mx, x a, and the #-axis,when from varies the the of distance the as point origin. square Also the moment of inertia about the #-axis. =
=
168. For of
Polar
Curve.
polarcurves
it is desirable to
second-order Let OP, OQ be two area
curve
a
r=/(0);
Ox
Element.
Second-order use
for
our
element
infinitesimal of different form. contiguousradii vectores of the
the initial line.
Let
0, 9 + SO be
INTEGRAL
202
the
CALCULUS.
angular coordinates arcs RU, ST, with and let SO respectively, the first order. area
a
RSTU=
to this order
of rectangle
Q.
and
Draw
cular cir-
two
0 and radii r and r + Sr of Sr be small quantities
centre
and
Then sector ORU
sector OST-
=
and
of P
r
"9 Sr to the second
RSTU
sides Sr
(RS)
be considered
therefore
may
and
rSO
order,
(arcRU\
Fig,48.
Thus
if the
at each pointR(r,9) is surface-density is (tosecondcr "f)(r, 9),the mass of the element RSTU order quantities the mass of the sector err S9 Sr, and =
is therefore
Ltdr==Q[2o-rSr]S9, the
summation
being for
from
all elements
r
=
0
to
r=f(9),i.e.
"rrdr\80, Q/(0) -i
in which
to be
and
the
9 is integration taking the limit of
infinitesimal values
regardedas constant,
sum
of S9 between
of the any
sectors
for
radii specified
CALCULUS.
INTEGRAL
204
0
=2/
(Art.164)
or
169.
Centroid.
C~% rZacoaO
/
pr*dOdr
Polars.
distance of the centroid of line may be found as before
The
sectorial
a
area
from
by findingthe sum any of the moments of the elementary masses about that line and dividing of the masses. by the sum Thus its
err
SO Sr
its moment abscissa, r cos
of
element
beingthe
the
6
SO ST.
a-r
r cos
0
y-axisis
about .
and
mass
rcos(9. a-rdOdr JJ
Thus
x
j\o-rdOdr r
and
similarly
o-rdOdr
sin 0
'~fl a-r
dO dr
half of the the centroid of the upper of Art. 168. example established the result for that semi-circle that We Ex.
1.
Find
circle in
the
Also
between
/*frcos
the
limits
^or c^^ dr=
r=0
and
r
=
0 for r, and 0=0
2acos
Tfji 0R" cos
53
"S
^^
15
'
to
INTEGRATION.
DOUBLE
and
I
I jnsin0
/rsm6crrd6dr=
J J
dO
-
L4
JQ
205
Jo
/~2
sin 0 cos40 dO
Ex.
Find
2.
cardioide
the
centroid
of
the
area
bounded
by
the
uniform.
being r=a(l+cos #),the surface-density
Fig. 50. centroid is The abscissa we have
evidentlyon
the
initial line.
To
find
its
/ Ir cos 0 rdOdr "x"
the
limits for
from The
0 to
TT
r
(and
numerator
being
from
'rdOdr r=0
double, to take =2
r=a(l+cos
0),and
for 0
half).
'"
fT-1 J
-
to
in the lower
cos0o?0
L3J0
cos20 0 1a3 /"'(cos + 3
+ 3 cos30 +
cos*0)dO
CALCULUS.
INTEGRAL
206
=
| T(3cos2"9 a3
cos46
+
1 ^4--
; '
'
'
4
o3;r
2
2
5
5
'
-
4
2
'
-
2
r~r2" Ia(l+cos 6)
/TT
dO
L2Jo
0
T(l+2
a2
=
cos
0+cos20)dO
0
rf r
2a2/ (l+ cos26")d"9
Hence
x
=
-?ra3
/-?ra2
=
-a.
varies as the nfh power Ex. 3. In a circle the surface-density 0 of the distance from a point on the circumference. Find the about an axis through 0 perpenof inertia of the area moment dicular to the planeof the circle. the Here, taking0 for originand the diameter for initial line, radius. is the r=2acos a The density 6, being bounding curve
=p,r". Hence moment
Hence
of the element and its rSOSr is //,rn+1S$Sr, axis is //,rn+38$ of inertia about the specified 8r. of inertia of the disc is the moment the
mass
f ffj where
the limits for
r
0 to 2a
are
cos
0, and
for
0, 0
double). Thus
Mom.
Inertia
=
rcos"+4(9 dO J?^L(2a)"+4 4
n
+
J
to
~
(and
INTEGRATION.
DOUBLE
Again, the
of the disc is
mass
/*2acos0
r"5"
2|J ^o
=
frcosw+20d0. _?^L(2a)w+2
=
+ 2
n
Hence
207
Inertia
Mom.
Jn
4
=
EXAMPLES. 1. Find
(a) when (ft)when
of
of the sector
the centroid
circle
a
is uniform, surface-density
the
varies surface-density
the
distance
the
as
from
the centre. the centroid of a circle whose the nth power of the distance from a
varies surface-density
2. Find as
Also
point 0
on
the circumference.
of inertia
its moments
(1) about (2)about
the tangent at 0, the diameter through 0.
of uniform of inertia of the triangle that the moment the a nd lines bounded the ?/-axis by surface-density mlx+cl^ y about is the #-axis, y=mtfc + c"ft 3. Show
=
Ml 6
where
M
4. Find
is the the
the coordinate
; and
axes
triangle,they are the placedat the mid-pointsof
bounded
that
the
2
2
respectively ---
and perpendicularto of the ellipse.
'
4
~,
its
as
triangleof
that
if M
those
of
be the
uniform
of
mass
equal masses
"
the sides. of inertia
1 about
and
4
show
same
moments
_--f fC" a2 62
by
m2J
triangle.
the
5. Show
"
of inertia of the by the lines
moments
bounded surface-density
about
\ml
of the
mass
V
GI-%
and
the
of
a
major and
about
a 2
plane,M"
line
uniform minor
ellipse axes
through the
are
centre
7)2
I
M
"
,
being the
mass
INTEGRAL
208
6.
Find
the
assuming from
between
area
the
surface-density
a
the
CALCULUS.
pole,
varying
(1)
the
centroid,
(2)
the
moment
Find
of
(1)
the
for
the
r=2acos#;
r=a,
inversely
and
the
as
distance
find
inertia
perpendicular 7.
circles
about
the
to
coordinates
of
the
through
pole
plane.
included
area
line
a
between
its
the
centroid
curves
(assuming
uniform
a
surface-density), (2) (3)
8.
about
the the
Find a
line for
(2)
for
pole
the
#-axis,
this
inertia
of
the
about
revolves
area
lemniscate
perpendicular
r2
its
to
the
plane
surface-density,
surface-density
distance 9.
the
uniform a
when
of
moment
through a
about
inertia
formed
volume
the
(1)
of
moment
from
varying
the
the
as
of
square
the
pole.
Find
(1)
the
coordinates
of
the
#=a(0-hsin$), (2)
the
volume
formed
of
centroid
y=a(l by
its
"
the
area
cos0)
revolution
(a)
about
the
base
(y=2a),
8
about
the
axis
(#=0),
about
the
tangent
at
the
vertex.
;
of
the
cycloid
ELEMENTARY
DIFFERENTIAL
EQUATIONS.
E.
I.
C,
DIFFERENTIAL
212
EQUATIONS. for the
and
definite value, the same different for different curves
of the
same
but
curve
family.
Problems
in which it is necessary occur frequently to treat the whole family of curves as, for together, each instance,in findinganother family of curves,
member set at
of which a
that
letter
ought
a
functions one
rightangle. And it will be the particularizing operations,
givenangle, say
manifest
to be
of the former
intersects each member a
for such to
not
as
appear
operatedupon,
individual
or
of the
curve
should
we
in the
constant
a
system
be treating
instead
of the
whole
familycollectively. Now be got rid of thus : a may Solve for a ; we then put the equationinto "
0(",y)
=
and
a,
the form
(2)
........................
differentiation with
regard to x, a goes out, and an equation involving x, y and yv replaces equation(1). This is then the differential equation to the family of curves, of which equation(1)is the typical equation of
upon
member.
a
In the formation
be
of the differential solve
to impracticable
case
we
differentiate the
for the
Q
=
and thus
respectto
then
x
obtaininga
example,consider values to giving special For
between
a
(1)
'
........
.............
equations(1) and (3),
relation between
is true for the whole
this
obtain
and
eliminate
In
constant.
equation
f(x,y,a) with
equationit may
x, y, and
yv
which
family. the the
family of straightlines obtained by arbitraryconstant in the equation
ORDER
Solvingfor and
OF
EQ UA TION.
AN
213
w,
differentiating, Blether-wise, first solving for m, we have
or
y=
without
yi and
therefore
This
then
=
m,
y=%yiis the
differential
passingthrough the origin,and
equationof
all
straightlines
the obvious geometrical that fact that the direction of the straightline is the same as line. of the vector from the originat all pointsof the same
172. Again, suppose the the familyof curves to be
expresses
representative equationof
ftx9y,a,b) 0, arbitraryconstants a,
(1)
=
b whose values containingtwo of the family. A the several members particularize singledifferentiation with regard to x will result in a relation connecting x, y, yv a, b ; say
4(x9y9yl9a9b)0 differentiate againwith regardto
(2)
=
If
we
obtain
a
relation
connectingx,
\MX and
will
the
V" 2/i"2/2"a"
x
we
shall
y, yv y2, a, b ; say
") "" =
(3)
cally equationsa and b may theoretiappeared be eliminated (if they have not alreadydisand there by the process of differentiation), result a relation connecting x, y, yv y2 ; say
from
these three
="" F(x"y,yi"yz) differential equationof the family.
173. Order
of
an
Equation.
We define the order of a differential equationto be the order of the highestdifferential coefficientoccurring in it ; and
we
have
seen
that
if
an
equationbetween
DIFFERENTIAL
214
EQUATIONS.
unknowns contains one arbitraryconstant the result of eliminating that constant is a differential equation of the firstorder; and if it contain two two
the result is
constants arbitrary
of the second that
so
differential equation
: argument is general arbitraryconstants we shall
n
our
and the result is proceedto n differentiations, an(l differential equationconnecting x, y, yv ...,2/n"
have a
And
order. to eliminate
a
to
is therefore of the nth Ex.
1.
Eliminate
a
and
order. c
from
the
equationx2+y2=2ax +
x -f yy^ a. Differentiating, Differentiating again,l+^+y^^^ and the constants having disappearedwe
c.
=
eliminant
have
obtained
as
their
differential equation of the second order (?/2 being differential w hich the highest coefficient involved), belongsto all a
circles whose
centres
lie on
the #-axis.
the differential equationof all central conies Form whose coincide with the axes of coordinates. axes the typicalequation of a member of this family of Here Ex.
and
we
2.
have
and whence is the differential
x(y? + yy2)-yyl=0 equationsought.
174. Elimination
an
irreversible process.
Now
this process of elimination is not in generala reversible process, and when wish to discover the we of a family of curves typicalequationof a member when
a
the differential equationis given,we are pelled comof integration, to fall back, as in the case upon
set of standard
which We
are
cases,
and
many
equations may
arise
not solvable at all.
however, that in attemptingto solve may infer, differential equation of the nth order we to a are search for an algebraical relation between x, y, and n
VARIABLES
SEPARABLE.
215
these constants arbitraryconstants, such that when eliminated the given differential equation will are result. Such is regarded as the solution most a general solution obtainable. DIFFERENTIAL There
175. CASE All
EQUATIONS
equationsin
other,
x's to
one
under
come
FIRST
THE
five standard
are
I. Variables
all the
OF
ORDER.
forms.
Separable. it is
possibleto get dx and side,and dy and all the y's to the this head, and solve immediately which
by integration. Thus
Ex.1,
if
sec
x-",
sec
y=
dx have
we
cos
and a
sin
integrating, relation containingone Ex.
"
x
=
cos
y
sin y + A
x =
(x
have
+
x
=
containingone
32
+
"
(y2+ y)dy,
J
2
therefore
A.
dx
)dx
-
\
and
,
xy-^*
y+l we
dy,
arbitraryconstant
If
2.
dx
x
logx"^
+y~+ A,
2
32
arbitraryconstant
A.
EXAMPLES. Solve
y
the
followingdifferential equations :
"
1.
I
dy=x*+x+\
dy
'
dx
/
4. Show
cuts
every
*++l' that member
y*+y+l
'
dxx*+x+l every
member
of the
of the set in Ex.
family of curves right angles.
2 at
in Ex.
3
DIFFERENTIAL
216 7. Show
equal
that all the
to
EQUATIONS.
for which of the radius
the square
curves
square
vector
of the normal is either circles or
are
hyperbolae. rectangular 8. Show
makes
that
a
constant
a
for which the tangent at each point angle (a)with the radius vector can belong to curve
r=Ae^
other class than
no
9. Find
cot
a.
the
for which equationsof the curves (1) the Cartesian subtangentis constant,
(2) the
Cartesian subnormal is constant, the Polar is (3) subtangent constant, the is constant. Polar subnormal (4) 10. Find
tangent
the Cartesian equationof the is of constant length.
176. CASE
II. Linear
P, Q,
.
.
.
,
K, R
the
Equations.
[DEF. An equationof when
for which
curve
the form
functions
are
of
x
constants
or
is
lies in the fact that said to be linear. Its peculiarity differential coefficient occurs raised to a power no higherthan the first.] As we are now equationsof the first discussing
order,we
limited for the
are
that
throughoutby multiplied
we
er
case
it will be
write it
may
d ,
dPe
/Pefccv n
)="^fPdx
"
yefPdx=\Qe/Pdxdx+
Thus a
the
' *
If this be seen
presentto
relation
between
x
differential equation,and It is therefore constant. The
factor
of the called
"
an
'
e
the given satisfying containingan arbitrary the solution required.
and
y
the left-hand member perfectdifferential coefficient is
which
rendered
equationa factor." integrating
Ex.
Integrateyi+xy
1 .
fxd~
e-**
Here
217
EQUATIONS.
LINEAR
x.
=
-
e2 is
or
integratingfactor,and
an
the
equation
be written
mav
d
-
-
(ye*)=xe*, ax *2
"
ye*=e*+A,
or
+
i.e.
y
Ex.
Integrate
2.
Here
the
=
^l +
-y=x2.
dx
x
factor integrating
2.
l+Ae
is e
Jldx =elogx=x,and x
the
equation
be written
may
*JH"-+ x*
and
xy=--+A,
177.
Equations reducible
equations,if
Many
not
to
linear
form.
immediatelyof
the
linear
form
_
be immediatelyreduced may variables. One
of the
most
to it
important cases
equation
y-n
Or
Putting we
have
yl~n
=
y-^dy=^
z,
by change of is that
the
of the
DIFFERENTIAL
218
EQUATIONS.
or
is linear,and
which
(l-ri)fpdx
,~
l-n
(\-ri)fPdx
,.,
=(1
e
y
\fr\ Q-~n)fPdxJ
=(1" ft) \Qe
ze
i.e.
its solution is
x-,
dx+A. A
Integrate-^ + ^=?/2.
Ex.1,
^-2^+^1;
Here
=
putting
or
f^ (l-ri))
v
We ri)
"
dx+A,
-=0, t7
dx and
the
factor being integrating -fix*
ej*
=e-loex
^(^=-1,
have
we
x
dx\x)
x
?=logi
i.e.
X
i.e.
X
-=Ax y
"
+ x sin 2y Integratethe equation-jf.
Ex. 2.
=
Cv"^7
Dividingby cos2ywe
have
ec2y-^ + 2#
tan
y"x^.
dx
Putting we
tany=2,
^
have
+
2^=^,
GW?
and
the
factor integrating
is
"J2xdxOr e*z, giving
DIFFERENTIAL
220
Find
16.
the
of
sum
the
nth 17.
the
polar
the
radius of
power
the
as
belong
to
the
and
vector
radius
the
of
whose
the the
family polar
of
for
curves
subnormal
varies
for the
which
the
perpendicular
pedal
equation
is
radius
of
curvature
the
upon
r2-p2=^
+ "^
Jc
being 18.
a
given
Integrate
constant
the
which
vector.
curves
square class
of
equation
the
that
Show
varies
EQUATIONS.
and
equations
A
arbitrary.
%* *
normal
as
XIV.
CHAPTER
OF
EQUATIONS
HOMOGENEOUS
CASE
178.
Equations
III
ORDER"
FIRST
THE
EQUATIONS.
ONE
CLAIRAUT'S
FORM.
LETTER
ABSENT.
Equations.
Homogeneous in
homogeneous
CONTINUED.
and
x
y
may
be
written
dx
(a)
In
this
a
result
obtain
case
of
the
y
obtain
v
+
x^
dv
comes
the
variables under
Case
are
for
-^, and
vxy
=
"j"(v)"
=
_dx
~"p(v)"v and
possible
form
Putting we
if
solve
we
'
x
separated
I., giving log
result
as
r 1
Ax "
and
the
solution
thus
DIFFERENTIAL
222
if it be
(6) But solve
for
-". we
EQUATIONS.
inconvenient
solve for
or
and
"
to impracticable
write
.
dx
x
p
for
and
"-
dx
.
have
we
y
x"f"(p)
=
...............................
with respectto Differentiating
(1)
x,
dx="["'(p)dp
or
x
-"
this equationwe have x expressedas a Integrating function of p and an arbitrary constant (2) Ax=F(p)(**y) Eliminatingp between equations(1) and (2) we obtain the solution required. .........................
Ex.
1.
Solve
(x*+y*)ty-=xy. dx
and
putting
y=v%,
^+v dx dv or
x"
=
-
dx
or
og=-2 a;2
Ay^eW.
or
Ex.
2.
Suppose the equation to x
dx
\dx)
be
'
HOMOGENEO
Then
p
US
=
(p +p2)
UA
EQ
+
x(l
+
TIONS.
223
2p),
p
log J,#+2logp
giving
-=0,
"
P
i.e.
and
the
jo-eliminant
between
p2+p=" x
\
Axp*="
and is the
solution
sought.
This
eliminant
But
when
is
it of
elimination
"
algebraically impossible to if performed, the when,
is or
p,
to leave manifestly unwieldy, it is customary to regard them containing p unaltered, and would equations whose jo-eliminant if found
solution.
EXAMPLES. Solve
the
differential
.=.
.
dx
2.
equations
x+y
the
perform result two
will
the be
equations
as
simultaneous
be
the
required
DIFFERENTIAL
224
179. A The
EQUATIONS.
Special Case.
equation
^-
-
~
is readilyreduced
-
"f
"
,
dx ax+oy+c the homogeneousform thus : Put x
to
"
bk +
a^+ by+ (ah+
^
TVi
c)
"
'
dg-a'g+b'' Now
choose h, k
that
so
.1
^e.
so
that
=
"
r-/ oc
"be^-
-
"
-"
,-
ca
^
-^ ao
"ca
"
a
6
^=
Then
equation being homogeneous we may variables an(i ^ne are as separable put n~v^ This
now
before
shown. is 180. There cannot be chosen
however, in which above, viz.,when
one
case,
as
a
b
c
6'
c''
_ ~~
a' Now
let
and
=m
"
a
dy
Then
so
that
Tx= n -~
--
7
"
a "*
= -
-
V
dx
my
drj (am
-
/)
+
c
+
+ ad b)rj
+ 6c
_
dec"" and
mrj +
c
-
-, ,
,
".
.
-,
+bc (am-\-o)r)-\-ac
n.
h, k
HOMOGENEOUS
The
EQUATIONS.
225
the integration beingnow separated, be at once performed. may 181. One other case is worthy of notice,viz., ax + by+ c dy dx~ "bx + b'y+ c" when the coefficient of y in the numerator is equalto that of x in the denominator with the opposite sign. For then we may write the equationthus + c')dy (ax+ c)dx+ b(ydx+x dy) (b'y differential equation exact an being ; the integral ax2 + 2cx + 2bxy b'y2 + 2c'y + A A. beingthe arbitrary constant. variables
_
=
"
"
=
,
Integrate =y-
Ex.l.
dx
Put
Choose
#="+
k, y
h and
Icso
=
ri+
x+y-Z k, so
that
that =
then Now
then put 77=0(1,
_
~
l+v
"-
v+1
1)2-
l
where
"=#"1
and
v=^~ .
x"\ E. I. c.
p
'
DIFFERENTIAL
226
Ex. 2.
Integratef
*+* =
dx
.
x+y
\
"
.#+y=??, then .=
..
Let
EQUATIONS.
dx
and
1
??
-
if] 1
'
"
^
where
?7=
EXAMPLES.
Integratethe equations: dy _
dx
bx+ay-b ^
"
1*
8 9. Show
that
a
particle #,
y which
moves
so
that
~
will
always lie upon
10. Show
fUL '
\"
that
~)
a
conic section.
solutions
of the
generalhomogeneous
always represent
must
families
of
tion equa-
similar
dx)
curves.
11. Show
that solutions
of
/(-,-j-} J are
\X
of y and some power the typical equationof in x, y and some
a a
homogeneous in
singleconstant, and converselythat member power
x,
CLX
of a of one
family of
curves
constant,the
if
be homogeneous differential
DIFFERENTIAL
228
Differentiate with
x, i.e.the absent
regard to
The
P
and
dx
Thus
EQUATIONS.
x
After
is performedwe integration and this equationand 2/ "/"(p) between of the givenequationis obtained.
the
=
which
eliminate p the solution
absent.
183. y B.
letter.
absent from then takes the form
the differential
Suppose y
equation,
fj1J
Since
-^
=
ax
"
this may
be written
ax
dy dx'
and
therefore if y
variable the
regardedas the independent remarks foregoing apply to this case also. be
Thus dx
(i.)if
convenient result of the form
solve
we
dx
.
,
a5T*S dx 7 dy
then
and the
=
"7-^,
is integral
dx
for
-^-, ^
and
obtain
a
ONE
LETTER
if this solution
(ii.)But
ABSENT.
for
-7-
be inconvenient
solve for
x
obtain
and
yy x
=
or
dy
we impossible
form
229
where (j)(q)
with
q stands
regardto
for
a
result of the
/v"
Then
-j-
tiating differen-
y, the absent letter,
Thus
and After
the
integrationwe and equation and x "f)(q). equationis obtained.
the
=
The or
y
student
absent,we
should
note
solve for
between solution of the
eliminate
q
that in either case,
~
x
this
given absent
if possible. by preference
ax
But
when
inconvenient solve we or impossible for the remainingletter and differentiate with regard the absent letter to the absent one\ thus considering in either case the independentvariable. as Ex.
1.
this is
Integratethe equation
'
Here
dy and
#=
"
2
is the solution. Ex.2.
Solve dx
Then
where
=
\dx) x
q=.
dy
DIFFERENTIAL
230 Then
EQUATIONS. absent lettery,
with regardto the differentiating /, 1 \dq l~
*
and
#
and
the
^-eliminantbetween
this
is the solution
equationx=q+~
equation
and
the
original
required.
EXAMPLES. Solve 1.
the
dy =
equations:
I.
y +
5.
6.
-B
\dx)
4.
(2a^ + ^2)=a2+2a^7.
\dx)
Clairaut's
V.
Writing^9 for
with Differentiating
dx
Form,
^=
have
we
-~-
=A+".
8.
dx
184. CASE
dx
y=px+f(p) regardto x,
........................
(1)
dp
{x+f(p)}0,
or
whence
either
-^-=0
or
....................
cc+//(p) 0. =
CLOu
Now
-f
=
"
givesp
=
C
a,
constant.
(2)
CLAIRA
UTS
FORM.
231
is a solution of the given differential Cx-\-f(C) C. constant equationcontainingan arbitrary Again,if p be found as a function of x from the equation (3) "+/(,) 0, Thus
y
=
=
........................
equation(2)will
stillbe satisfied, and if this value of which is the same or p be substituted in equation(1), if p be eliminated between thing, equations(1)and (3) shall obtain a relation between we y and x which also satisfiesthe differential equation Now to eliminate p between
y=px+f(p)}
x+f(p)I
0=
is the
same
as
to eliminate
0
i.e.the
same =
(2) The
x+f(0)J
=
the process of findingthe envelopeof Cx+f(C) for different values of 0. therefore two classes of solutions, viz. : as
the line y There are
(1) The
0 between
linear
called the solution,
"
completeprimitive," constant. an containing arbitrary "
solution envelopeor singular containing and constant from derivable not no arbitrary the complete primitive by putting any numerical value for the constant particular "
in that solution.
The
geometrical relation between
these
two
tions solu-
is that of a familyof lines and their envelope. It is beyond the scope of this book to discuss fully the theory of singularsolutions,and the student is referred to largertreatises for further information upon
the
subject.
EQUATIONS.
DIFFERENTIAL
232
Solve
Ex.
y=jt
JP
and m
is completeprimitive
Clairaut's rule the
By
the
is the result of
solution envelopeor singular the above
between
eliminating
equationand
o=*--2. m2 y*="ax.
i.e. The
student
y2- 4a#
the
=
will at
equation to the
y=mx+"
once
well
a
recognizein and parabola,
the singularsolution in the completeprimitive
equation of
known
a
tangent
to
the
parabola. EXAMPLES. find the
and the complete primitive, down Write cases : solution in each of the following
envelope
"
4. y" 5. y
6.
185. The
(x
=
"
a)p
"
p^.
(y"
equation y=x"P(p)+*Kp),
(i)
.....................
with regard to x, by differentiating variable. then considering p as the independent have For differentiating, we
may and
be
solved
"
whence
-=
--
dp which
is
the solution being linear,
["P(P)*P r .,,Ji eJ"t*P)-P=_ (P) xe"t*P)-P=_
tW-Pdp+
A
.......
(2)
233
EXAMPLES.
If
equations(1)and originalequation
between
p be eliminated
now
(2),the completeprimitive of the will result. Ex.
Solve
We
y
have
=
2px+p2.
(1)
,
p
.
(2) p^x" %p3 A giving be found from these two equationsmay The jo-eliminant now in equation(2). by solvingequation(1)for p, and substituting But if it be an objectto present the result in rational form, we may proceedthus : 0,\ By equation(2) 2p3+ 3p2#+ SA from (1) 0. / + Ip^x-py j93 "
"
..............................
"
=
=
Hen
And
p*x Zpy
ce
-
-
3A
=
equationand
this
between by cross-multiplication
givingas
0.
the eliminant
4(y2+ 3Ax)(x*+y)
=
(xy
-
3 A)2.
algebraic p being process of eliminating the equations(1) difficultor impossible, in many cases and (2)are often regardedas simultaneous equations but the is the solution in question whose ""-eliminant actual elimination not performed. 186. The
EXAMPLES. Solve the
equations:
1. y= ,
2. y
=
3. y= 4. y=
p
DIFFERENTIAL
234
The
8.
and
T,
OT2
PTto
is
the
the
differential that
the
the
in
the
Form
length
A
12.
is
for
and
tangent
the
as
the
as
possess
the
by
Obtain
when
which
the is
tangent
a
complete tion solu-
singular
of
area
the
of
and
which
for
curves
the,
between
intercepted
tangent Obtain
triangle
constant.
equation the
the
interpret
the
complete
solution.
singular
satisfies
that^"=0
made
of
1888.]
which
curves
intercepts
tangent,
constant.
the
curve
all
constant.
the
of
portion
and
the
differential
axes
primitive
of
and
axes
the
coordinate
is
curves
the
of
[OXFOKD, of
differential
the
x=\
;
determine
its
and
y=p\x"p))
equation
equation. 1889.]
[OXFOKD, IS.
Find
the
in
Oy
inclination
the
of
tangent
axis
question.
the
by
11.
sum
the
meets
curve
curve.
of
axes
curves
bounded
the
a
equation
equation
Obtain
10.
the
the
to
Find
coordinate
primitive
also
proportional
of
P
point
any
Ox.
property the
on
at
axis
Find
9.
the
tangent
EQUATIONS.
complete
primitive
and
singular
solution
of
the
equation
14.
is
Show
reduced Hence
solution.
that
to
write
one
by
putting
of
Clairaut's
down
Interpret
its the
x2"s
and
y2
=
[OXFORD, ", the
1890.]
equation
form.
complete result.
'
V^yj
\
dx
primitive
and
find
its
singular
DIFFERENTIAL
236 Then
yi
=
2/2
=
Thus
on
But
EQUATIONS.
zj(x)+zf(x); zj(x)+^'(x)+zf"(x).
substitution
get
we
+ Pf(x)+ Qf(x) /'(*) =
0
by hypothesis.Hence
z
an
equationwhich is linear factor is The integrating
for zv
or
is and the first integral
the second integral whence may effected. solution and the Ex.
Solve
y=x
once
L
da? Here
be at
makes
dx
-r^
Put then
Hence
**
/K" 4- 3^
and
the
factor integrating
is
e^
x
or
x*e
4 .
obtained
237
x*
j
Thus
EQUATIONS.
ORDER
SECOND
~(zlx^)=x^ a*
5
z1x2e*=~+A
and
_*
whence and
z=-\e
the solution
r
+ A
*
,
\ "-e
J a?
requiredis -=
5
189. CASE A. If
x
II
be
letter absent.
One
absent, let y1=p, -
and
the
=
"f"(y, p, p-S-\Q,
takes the form
=
dy/
\
and
0
"f"(y, yv y2)
equation
is of the first order.
B. If y be the letter
absent,let
yl =p,
*-" and
"f"(x, yv y%) becomes
and
again is of
Ex.1. Here
The
or
Solve x
the firstorder.
the
is absent.
equationyy2+#i2=2#2. So
factor integrating
puttingy"=p
is e^
p2y2"y* +
vAy or
constant
and
y2=p^?, have
y2,
=?/4+ a4,say.
we
EQUATIONS.
DIFFERENTIAL
238
Hence
or
+ A). y* a2sinh(2#
i.e.
=
l+#i2=#y2#i"
Solve
Ex. 2.
y is absent.
Here
So
puttingy-^"p^
dx or
pdp
-"--"5,
=
"
1+jtr
x
i.e.
logx
logVl+p2
=
+ constant.
^.e.
ady" ^Jx*
or
"
and
dx,
oy=i?^2_"
giving a
a?
b
constants. being arbitrary
EXAMPLES. Solve the 1. ^2
=
followingequations: "
6.
1.
2.
1+3^=^2.
7.
3.
i+y!2-^2-
8-
1
y2+"/i-y=-e
_L2 2-
^"/
4.
5. "3/2
=
10. Solve
that
a
9y22=4iyi-
^
=
y#2=#i3-#i-
[OXFORD, 1889.]
(l+.yM the
0 when
=2/" havinggiven ~#2)^~#(^)
equation(1 y
=
0.
[OXFORD, 1890. ]
ow?
11. Given
find the
that #2 is
a
value
completesolution.
of y which
satisfies the
equation
[L 0. S., 1894.]
REMOVAL
Let
Linear
General
190. Term.
Q Pv P2, Puttingy vz, we
where
.
.
.
239
Removal
of
a
generalequation
more
of
given functions
are
,
x.
have
=
y2
TERM.
A
Equation.
consider the
next
us
OF
vz2 +
=
2V"
+ v2z, etc.,
whence
n(n"
1)
, -
--
-
2
The coefficientof 0n_i is If then v be chosen so that P
dv
"
or
"
v
the term
chosen
the term containing 0W_2 The coefficient of z is
if
a
will make
zl the =
value this
=
e
of
v
been
will have
zn-i involving if be so Similarly, v equation
and
v
n
as
to
will have
the satisfy
been
be found
can
removed.
expressionvanish,we
or
ential differ-
removed.
guessedwhich can, by writing
therefore z2 zn rjn-i, reduce rjly etc.,and degree of the equationby unity. The student
rj,
should
and
notice
=
=
that
this
expressionis
the
same
in
DIFFERENTIAL
240
form
the
left hand
if any
solution
as
Hence
EQUATIONS. member y
"
v
of the
given equation. be found or guessed of is right hand member
can
the
the given equation when omitted, we by writing y can, vz, reduce the degree of the equation.
and
=
Canonical
191.
In the
the
substitution
will
by what equation to
But at
equation
y
has the
"EXACT"
p
of the
above
can
For
be
"
q.
denoting
by
of this
equation
given
has
yq,
=
not
been
EQUATION. is
xp-r~
etc.,
Thus
the
simpler form
integratedwhatever
\xPyqdx
Y\,
degree
reduce
stated
DIFFERENTIAL
is
second
an
exact
^
and
"
e~l ^dxz
sometimes
general solution present effected.
When
=
been
the
192.
Z^
Form.
of the
case
then
y may
be.
differential.
EQUATION.
DIFFERENTIAL
EXACT
It will be noticed
that when be effected.
cannot
By aid of quicklywhether 193.
the above
given
a
lemma
"
or
q =p
241
the integrati
p
may often see equation is " exact/' For we
if all terms of the form xpyq in which p is first removed, we tell at once can frequently whether the remainder coefficient or not. Ex.
is a
q be spectio by in-
"
differential perfect
#2
Here, by the lemma, #2y5 and
x?y" are
perfectdifferential
and obviouslyocy^-\-yis coefficients, of xy.
Hence
first
a
the differential coefficient of integral this differential equation is
obviously cos
=-
194. A A
more
generaltest
equationmay
Test. for
an
"exact"
be established in the
For upon
n-
differential
generalcase
forms
the coefficients P0,Pv have, providedthey be functions of x. whatever
.
General
more
x+A.
...
,
Pn, V may
denotingdifferentiations by dashes, we integration by parts
Bysdx=Pn-
32/2 -
Pn
-
8^1 +
P"n
-
zV
-
IP"'n
etc.
Hence
p. i. c.
upon
addition it is obvious
have
that if
-
EQUATIONS.
DIFFERENTIAL
242
given equation
the
exact; and
is
tegral its first in-
that
is
Is the
Ex.
and
=
PQ'" 24^ P3 P2'+ PI equationis exact ; and its
the
Thus
=
-
-
exact
x
?
have
the test,we
Applying
sin
+ 1 2x?y2 + SG^2^ + 24#?/ equationcfiyz
-
72^ + 72^ first
24^
-
0,
=
integralis
or
This
which
be
again will is satisfied.
perfectdifferential
a
Hence
a
(8#3 43%
second
-f ^4yx =
-
-
if
integralwill sin
x
+ Ax
4^73y+^4y1= sin^+^^
or
"
again be tested. may is third and final integral
which
^
But
it is
now
be
+
B,
+
J?,
obvious
that
the
cos.*?+
=
IB
EXAMPLES. 1. Show
exact, and 2. Solve
^7/3+ 3. Write
that
the
solve it the
equation
completely.
equation
.
+ sin
6^/2 +
%i
down
first
x(y* %i) + ~
of integrals
the
cos
X33/2 !/} =
-
sin
^
followingequations :"
'(a) (b) (c) 4. Show an
that
if the
integratingfactor
equation
equation P2y + P^y^ + P0"y2 //, will satisfy the //,,then =
F admits
of
differential
DIFFERENTIAL
244
EQUATIONS.
Hence
is
solution of
equation(1) containingn arbitrary constants,and is therefore the most generalsolution to be expected. No more generalsolution has been found. The portionf(x) is termed the Particular Integral the n arbitrary and the remainingpartcontaining (P.I.), constants,which is the solution when the right-hand member of the equationis replaced by zero, is called the Complementary Function (C.F.).If these two partscan be found the whole solution can be at once a
written
down
as
their
sum.
remarkable
196. Two
Cases.
in which readilyobtained. generally There
are
(1) When
two
the
cases
these solutions
Pv P2, quantities
Pn
...,
be
can
all
are
constants.
(2) When
the
equationtakes
r7n-2n/
Jn-l
ri
the form '
F"
2+-'-+^=
av a2, ..., an beingconstants and V any function of x. is readilyreducible, The solution of the second case as will be shown, to the solution of an equationcoming under
the firsthead.
EQUATION
WITH
COEFFICIENTS
CONSTANT
MENTARY COMPLE-
"
FUNCTION. 197. Let us therefore firstdetermine such an equationas + a22/n-2+... + 2/n+^i2/u-i the coefficients being constants
confine "
our
attention
to
ComplementaryFunction
the "
the solution of
^n2/
=
0,
; i.e.for the
presentwe
determination
in the first case.
(1)
.........
of
the
Y
COMPLEMENTAR
As
mn
+
245
put y Aemx, and we have a1mn-1+ a2raw-2+...+an 0
trial solution
a
FUNCTION.
=
(2)
=
.......
equationbe
Let the roots of this
774, m2, ms,
...,
mn,
then supposed(forthe present)all different,
and all solutions,
are
therefore also
+...+A (3) nem*x, A^x + A2e^x+ A3em*x is a solution containing n arbitraryconstants Av A2, An, and is the most generalto be expected. A3, y
=
......
...,
198. Two If two
Roots
Equal.
of
become
equal,say the of first the solution (3)become two terms mx m2, be regardedas a and since A^ + A^ may (A!+ J.2)e"11*, roots
equation (2)
=
singleconstant, there unity in the number (3) is no longer the expected. Let
us
this
examine
Put
is of
apparent diminution by arbitraryconstants,so that most general solution to be an
closely.
more
m1+A. ^x + A 2e(TOi+*X" 97i2 =
A
Then
h?x2
r =
~~\
Alem^x-\-A2(^x\ l+hx+-^-
+
...
rhy?
=
(Al+ A2)^x
Now
A^ and and quantities, of two
A2 we
other
+
AJi
are
may
.
~\
I. xem^x+Azhem^~^
two
+
.
..
independentarbitrary
therefore
express
them
in
independentarbitraryquantities by two relations chosen at our pleasure. First we will choose A2 so largethat ultimately small may be written "2, A2h when h is indefinitely an arbitraryfinite constant.
terms
DIFFERENTIAL
246
EQ
UA
TIONS.
will choose A1 so largeand of opposite we Secondly, signto A2 that A^+A2 may be regardedas an arbitrary finite constant Bv Then the terms
vanish with h since Aji has been considered ultimately is confinite and the expressionin square brackets vergent and Thus
h
contains
as
factor.
a
A^^+A^e11^ may, when m2 mv and therefore ultimatelyreplacedby B1emiX+B2xemiX" in the whole of arbitrary constants the number
be
the terms
=
solution remains we n, and the generalsolution in this 199. Three
Consider
therefore
obtained
have
case.
Equal Roots. the become
next
three of the roots of The equal,viz.,m1 m2 m3.
case
when
equation(2) have already been terms, Alem"+A#m*x+A2f?ri*x, by (Bi+B^e^+Atf"**. Let
ms
=
A^x
+ Ae'W
for A+Aw
Thus
and
AjPtifc A^x( =
=
we
may
so
choose
Ov C2,03 being any
placed re-
mx + h fcZftZ
/
Then
=
=
1 + kx + we
-^-
\
+...)"
have
AB,52,and Bv
that
arbitraryconstants, whatever
k
COMPLEMENTARY
FUNCTION.
be, providedit be
may
finite
absolute
not
247 But
zero.
AJc2
series within being chosen a the square brackets beingconvergent, it is clear that when k is indefinitely diminished, the ultimately, form limiting
of this
200.
Several
In
similar
a
roots of the
and quantity,
is expression
Roots
Equal. it will be
manner
equation(2)become m^ m2 loss of =
there
= .
.
that
if p
equal,viz., =
.
obvious
mp,
in our solution if generality expression + KjxP *)"*", (%!+ K^x + Kfl?+ portionof the complementary corresponding
will be no substitute the
we
the
-
.
for the
.
.
function,viz.,
A^x
+
+...+ Apew*"x. A2em*x
201. Generalization. More
if generally,
be the
ential complementaryfunction of any linear differequationwith or without constant coefficients, what is to replace this expression to re-tain the so as when generality mx m2 ? =
Let
m2
Then
and the terms
become + A2"p(m2) Al"f"(m1) h2
Now two
putting A^A^ Bly AJi "2) arbitraryfinite constants,the remainingterms =
=
DIFFERENTIAL
248
EQUATIONS.
ultimately disappearwhen we approachthe limit diminished. which h is indefinitely Thus Al"f)(ml)+ J.20(m2)may be replaced by
in
number (n) of arbitrary retainingthe same constants An B1952,A2, A^ in the complementary function as it originally possessed. And as in Art. 200 we proceedto show that if may viz. ^i1 m2= p roots become equal, =mp, the terms + 420(m2)+ +Ap"j"(mp) J.10(ra1) be replaced by may thus
...,
=
...
"
when
the
of generality
.
.
the solution will be retained.
results of Arts. 198, 199, 200 are of course ticular parof this,the form of "p(/m^) emix. cases being
The
202.
Imaginary Roots.
When
a
root of
equation(2)of
Art. 197 is imaginary,
it is to be remembered that for equationswith coefficients imaginaryroots occur in pairs.
Suppose,for instance,we where
i
Then
=
\/
"
have
1.
the terms
A^+A^e"** may
real
be thrown
into
real form
a
(Al +
thus
: "
bx bx)+ A2eax(cos bx bx + (A1 ^2"6a*sin A2)eaxeos 1
=
A^
or
sin
"
-
sm
bx,
i
sin
bx)
FUNCTION.
COMPLEMENTARY
where
249
constants Bl and B2 replace arbitrary (Al A2)irespectively. then cos a, B% p sin a,
the two
A^+ A2 and Let B^ p =
"
=
JB*
=
Then
+
"22
and
-
ijgF.
"2sinbx p cos(bx a). We may thus further replace bx by CLeaa!cos(6aj jB^cos bx + B2eaxsin constants. where C^ and 02 are arbitrary ^cos
203.
fr#+
tan
=
a
=
Repeated Imaginary
"
Roots.
For
repeatedimaginary roots we may proceedas that when before,for it has been shown 7772 ??i1, Alem^x+A^x may be replacedby (J^+J?^***, and if m4 by m3, A^X+A^X may be replaced =
=
If then mx
=
m2
=
a
+ ib and
m3
"
m4
=
a
"
f
6,we
may
replace by that is
(
by
sin te] bx + (Bl- B3)i + 53)cos eax[(Bl + xeax[(B2 + ^4)cos60?+ (B2 "4)^sin 6 and therefore by 6^+ (72sin 6x+ (74sin +cceaaj((73cos e^CC^cos 6aj) that is by 6aj+ ^ + cc(73)cos tf*(Ci which is the same or thingby -
Any
of the last three
constants constants
which
forms
replacethe Av A2, A^ A^ and
contain
four
originalfour thus
arbitrary arbitrary
retain intact
the
DIFFERENTIAL
250
EQUATIONS.
(n) of arbitraryconstants requisite proper number to make the whole solution the most generalto be expected. And this rule may obviouslybe extended to the case when of the imaginaryroots any number equal.
are
204. Ex.
1.
Solve
equation^-
the
dx
dx2
Here
whose
trial solution
our
roots
are
Accordinglyy
1 and
is y=Aemxy and
obtain
2.
A^e*and
"
we
and
y
=
A2e2xare
both
solutions, particular
y=A1e*+A2e2x
is the
constants. generalsolution containingtwo arbitrary
Ex.2.
Solve
-*V-a?y=0. aOC
Here the auxiliary equation is w2 and the generalsolution is
or
as
it may
be written y
by replacingAl by
"
a2
ax .Z^cosh
=
Bi+B*
and
+
^sinh
"a,
A2 by
ax
B^~B 2
the auxiliaryequationis m2-f-a2=0 the general solution is y
or, which
is its
Ex.4
Solve
=
=
stands
+
roots
m"
+ai.
^t2si
Bl
?-4| ax?
D
AjCosax
with
equivalent, y
where
m"
Solve
Ex.3.
Hence
roots
(ifdesired)
2
Here
0 with
=
for
ax
"
.
ax
+
5-2y ax
=
0
or
(D- l)2("-2)y 0, =
3-
4-
5.
6.
EQUATIONS.
DIFFERENTIAL
252
S-9 S~3 g=y. g=y.
9.
10.
("-
11. 12.
THE
of such
a2,
a
=
beingconstants,and Fany
an
...,
next
we
INTEGRAL.
Having considered the complementaryfunction an equationas F(D)y V where F(D) stands for
205.
av
PARTICULAR
turn
our
We
most
propose useful of the processes above
the
write
may
of x,
of
to the mode
attention
and particular integral,
and
function
obtaining to givethe ordinary adopted.
equation as
2/
=
1
where (or [/(D)]F),
satisfies
206. "Z""
is such
^7^
the
operatorthat
an
fundamental
laws
of
that
the
Algebra. It is shown
in
the
Differential
Calculus
satisfies (denoting -y- j
operatorD
(1) The
Distributive
(2) The
Commutative
i.e.
Law
of
Law
D(cu)
=
Algebra,viz. as
far
c(Du}.
as
stants, regards con-
INTEGRAL.
PARTICULAR
(3) The and
Index
253
Law, i.e.
integers. beingpositive Thus the symbol D satisfiesall the elementaryrules the with of algebraical of combination quantities with regard to exceptionthat it is not commutative m
n
variables. It therefore
identityhas a analogue. Thus
rational
algebraical correspondingsymbolicaloperative since by the binomial theorem
follows
that
any
'
7? l
+
(T)
~Li
"""
\JL
2i .
have
we
may
by
an
be inferred without
Let
proved in the positive integer,
a
us
define the
"
"
Differential Calculus that
D~r operation DrD~ru
Then
.
Operationf(D)eax.
It has been if r be
operatorswhich proof
further
JL
207.
for
analogoustheorem
=
to be such that
u.
D~l
and we shall representsan integration, D~lu no constants arbitrary suppose that in the operation added (forour is to obtain a are objectnow and not the most generalintegral). particular integral Now since Dra~reax eax DrD-reax,it follows that =
=
D~reax
=
a-reax.
Hence it is clear that Dneax aneax values of n positive or negative. =
for all
integral
DIFFERENTIAL
254
EQUATIONS.
pansion f(z) be any function of z capableof exof z, positive in integral or negative powers ^Arzr say, Ar beinga constant,independentof z).
208. Let
(
=
Then
result of the
The
be obtained Ex. 1.
f(D)eaxmay operation D by a. by replaci'ng
Obtain the value of
-
"
"
"
therefore
-e
^
the rule this is
Obviously by
"*
Obtain
the value
E,
2,
By
the rule this is
or
g.
of
e3a/'=-^-e
EXAMPLES. 1. Perform
the
indicated by operations
'
^
3.
Apply
Art. 208
to show
that
m^7 /(Z)2)sin mx /(Z)2)cos
=/( =/(
209.
Operation f(D)eaxX.
Next
let y
=
eaxY,where
"
"
Fis
m2)si m2)c
any
function
of
x.
INTEGRAL.
PARTICULAR
have
we
Dreax
since
Then
255
areax,
=
Leibnitz's Theorem
by
1D Y+ nC2D2Y+...+Dn F+ n(71an edx(an -
yn
=
which, by analogy with 206), may be written
the
Dneax Y= n
Binomial
Theorem
F), (Art,
eax(D+ a)nF,
beinga positive integer. Now that
so
X
let we
from
Then
write
may
above
DneaxY=eax(D+a)nY nX
Dneax(D+ a)
~
or
and
therefore
Hence
in all
=
eaxX,
D for
cases
integralvalues
of
n
or positive
negative DneaxX 210. As
in Art. 208
eax(D+ a)nX.
=
shall have
we
f(D)eaxX
=
be
That is,eax may the leftof the by D + a.
Ex. 2.
"
-
D2
"
"
"
transferred from the rightside to operator f(D) provided we replaceD
-
-
4D + 4
e2xsinx
=
e2x~ D
sin *
x"
"
e~xsin x.
DIFFERENTIAL
256
EQUATIONS.
EXAMPLES. 1. Perform
the
operations
1
(D
If*
-
1 l
h
(D-I)*6
'
D-l
Operation/(7"2)
211.
sn
D2
have
wwu
(
=
m2)y
-
cos
and
X'
that
2. Show
We
1
o
X
sin mx, cos
therefore
Hence,
before,Arts.
as
a^ j;/ r"9\
n
J
Ex.
D
^
sin
/v
) '
feaxsin bx dx
and
208
mx
y
"
cos
6^ (Art.210) eax(D+ a)~1sin
=
-_
mx.
"
Z)-1eaxsin 6^7
7)Wri
hv
f'Arf 91^ xxi li, Zi L 1
X/^iSJlllf"x/
sin ".#
a
sin
o\
f( m2)
=
cos
=
210, it will follow
6
"
cos
^
bx
tan-1- Y ^sin^.r
ea*(a?+62)
-
EXAMPLES. of integrals eaxcosbx, e^sin2^,e^sin3^,si the operations Perform
1. Find
2.
by
this method
-sin 3. Obtain
by
means
the
2^,
-_
sin 2^7.
values of the sine exponential mx. /(/")cos mx, /(Z")sin operations
of the
cosine the results of the
_"I L_cos,r,
and
y
INTEGRAL.
PARTICULAR
Operation
212. Let
257
consider the
next
us
operation
F(z)is a function of z capableof expansionin integral positive powers of z. Let F(D) be arrangedin powers of Z),then if no odd the result may be written down occur by the powers rule of Art. 211. foregoing where
Thus ^ sin S1T1 2#
""
sin %x"
=
"
But
if both
follows
proceed
as
and the
and
even
the odd
odd
Group and together,
"
powers
"
sin 2#.
51 occur
powers the
:
" ~
+ 16-64
L-4
we
may
powers together then we may write
even
operation sm
mx
./T^ox
^
.
mx
sin
=
/TV,V
^
^ "
Upon we
may
m2)sinmx
mx(
"
"
m2)cosmx
that in practice it will be seen examination write m2 for Z)2 immediatelyafter the step "
writingimmediately 1 "
E. I. C.
"
"\
f)~7
^ R
sin
mx"
EQUATIONS.
DIFFERENTIAL
258
"
mi
"
in/
1.
"
"
i
J-'Xv
in
i
Obtain
of
the value
.
"
"
"
,
ma;, etc.
4^0 sin
"
"
Ex.
)
SYT^ fjsrS
r-y^
or'
-
="
sin 2#.
^
Thisis sin D
"sin
2#, 2^,
-
J.O cos
2^
the value
of
^
or
Ex.
This
2.
Obtain
expression
^
"
-^e2*cos ^p.
"
^
e2*-^
=
sin 2^7.
"
"
cos
-
x
^
each [replacing
Z"2 by
1]
"
e2*
1
____
=
"
"(cos
x
"
sin
x).
4
EXAMPLES. 1. Perform
the
operationsindicated
in
the
pressions followingex-
:-
D
Z"3
-e*sin
x
+
EQUATIONS.
DIFFERENTIAL
260
EXAMPLES. the
Perform
operations
CD+l)(Z"
2)
+
2-
3.
214.
J?
-
Cases
COsh
COS
37
07.
Failure.
of
applying the above Particular Integral, cases with.
met
We
procedure 215.
Ex.
be
to
adopted
illustrate
in such
To
If
we
208, the
Art.
apply
Integral
Particular
the
obtain
result
i^i We
this
evade
may
operation by applying
210
'
have
we
becomes
or
""-
obtain
difficultyand Art.
when
we
the
the
result
of
the
have
particular integral required. another of substituting method, however, Instead,
which
of
course
is Ae*.
Function
Complementary
the
cases.
dx The
are
obtaining a frequently
(^L-y=ex.
equation
the
Solve
1.
failure
of to
propose
of
methods
In
is the
operation
Writing
=
x(\ +
"
-"
h)
more
instead
carefully. of #,
we
have
let
us
examine
263 OF
CASES
expressionthe portionLtex//ibecomes taken with the complementaryfunction Aex
infinite,
this
Of
be
may
FAILURE.
arbitrarywe
regard A
may
+
; and
arbitraryconstan
as
a
new
to contain
a
negativelyinfinite
-
b
A
/i
for
we
A
suppose
may
the term I/A. xe* is the Particular term
por
to cancel
The
remainingterms indefinitely.
the
Solve
Ex.2.
y
and
In
sin 2#.
211, we
We This
is
and
oo
2#.
cos
parts "i
of two
part, if
"
e*
apply the
we
o
ri
fail.
so
,
when limit,
the
consider
now
this second
2HL",i.e.
get
clearly
A sin Zx + B
=
integralconsists particular
The
Art.
^ Ct/X
equation
complementary function
The
h is decre;
is
solution
whole
when
vanish
contain h and
The
The
Integraldesired.
h
Q, of
=
--
sin2.i'(l-
"
expression 1
_1
=
4
i_(i+A)a 1
1
I
JT2^U^X
9A
COS
"^IX "*"COS
^^ S^n
^^J?)
-
1 sin 2.2? " "
l
fi
o
=
(a
"
term
1
which
may x
-
the whole
,
ofr A
4
function) Thus
2.27+ powers
cos
-x
solution
be included ^
c^s +
in the
(terms which
of the. differential
complemvanish
v
equation
g JOS
"
"
DIFFERENTIAL
260
:.
the
equation (D2+ 3D)(D -l)2y e* + complementary function
Solve
3.
EQUATIONS.
=
'ere the
integralconsists particular
e~x + sin
is
=
parts,viz.,
"?
x_
iy*6 ~(D-I*'
(a part going into +"
ex +
the
4- x1.
plainly
of four 1
1
x
=
*_ JL
T
'I?''
4~4
complementary function)
(termswhich
vanish
with
,^-PP,
10
3-Z)
1 sin
-
x=
6 + 2Z) =
(3 sin
#
"/
2(9 -
cos
-
^)/20.
jjinally
open
44
ie
y
the whole =
solution
is
Al + A2e~3x+ (A3 e2x
3 sin
x
"
cos
x
5#2
.3? .
+
~~
+'+""l"
44 ,
h)].
Ex.
4.
The
C.F.
To find the which
the
Solve
EXAMPLES.
TIVE
USTRA
ILL
263
equation"^-?/ ^sin^. =
is
P.I.
we
have
."-a
is the coefficient of
i
in 1
"?".in
#*,
l. "
"*"
plX
rp
-4^-6/^Tr. l **
Thus
the
l
~^l-^D..\^
COSJP
is
P.I.
_
8
and y
the whole =
A
^inh ^
solution +
'
3^ gin ^ 8
is
^2coshx
+
A3sm
x+A
x
4cos
x
+
c^s^' ^ -
sin
^p.
EXAMPLES. 1. Obtain
the Particular
0)
sina7'
7TTT
indicated by Integrals
(5" /n (6)
nT-T8*1111*'
_
(7) /na
w^ovn-Q-x*'(sinh#
^/
n
+ sin
^?).
^o,(^ COS
+ cosh
-
o
COS
6^).
.
o
DIFFERENTIAL
264 Solve
2.
EQUATIONS.
differential
the
(3)
+y
equations
=
Cfc#
(5) (Z)-
(4) (D*-l)(D*-l)y=xe*. (6) (ZP-3D2-3D
I)y
+
=
e-x
(8) (Z"2
(7) OD3-%=#sin.". (9) (Z"2 (10) (^"-
216.
The
Operator
""-. CvQC
reducing an
where the
equation of
Av A2,
coefficients
constants, to a form constants, arises from
this
case
are
=
-TT
therefore
e*,and
the
that
therefore
x~ax
-ji
putting
=
-^at
operators d
d are
for
stand
D
Let
equivalent.
-j-.
have
-*-
dx\
)X
n
_
n nf\n
Z. _
dxn
" "
(x \
all
et.
=
at
It is obvious
in which
are
...,
in
the class
x
In
peculiarservice
renders
which
transformation
A
__
dx
11
-I-1 ]X /
. ~
1 ___
dxn~l
x-j-
and
dx
Then
we
EXAMPLES.
Now
265
in succession
putting11
2, 3, 4, ...,
we
have
etc. ,
Hence
or
generally
the order of the operations reversing D(D-l\D-Z)...(D=
Ex.
the differential
Solve
Putting x
=
equation
c?,the equation becomes
D(D -l)(D- 2)y+ 2D(D
%
-
+
3%
-
3#
=
(
or
i.e.
(D
giving y
=
Ae* + B
-
cos
-"
,
~+;rIog EXAMPLES. Solve
the differential 1.
s 2
2.
equations
x-
dx
+ --^
3.
+
cfc1
a?
-^ + (^
=
+ x [log^-]2
3^++2/=^ ote2
rfa?
dx*
dx
4. "i
dx? 5.
'
sin
log^
+
sin q
loga?.
CHAPTER
ORTHOGONAL
XVII.
TEAJECTOEIES.
MISCELLANEOUS
EQUATIONS. ORTHOGONAL
TRAJECTORY.
217. Cartesians. of a equationf(x,y, a) 0 is representative The to family of curves. problem we now propose is that of findingthe equation of another investigate of which each each member cuts family of curves in of the former family at rightangles. And member such a problem as this it has been alreadypointedout The
that
=
it is necessary
to
treat
that so familycollectively, a ought not to appear in
It has may
been
shown
be eliminated
all members
of the
first
the
constant particularizing the equation of the family. in Art. 17 1, that the quantitya between the equations
.
*dx
Let this eliminant
This
is the
'dy dx
be
differential
equation of
the
first
family.
DIFFERENTIAL
268
EQUATIONS.
x+yJL=a,
Here
cLx
.v2+y2 2x( x -\-y" ),
and, eliminating #,
Hence
the
new
=
differential
^2+ 2^-^2
or
be
equationmust
o,
=
(3)
...........................
ay
homogeneous equation, and the variables separableby the assumptiony vx. as However, this being the same equation (2) with that x and y are interchanged, must its integral which
is
a
become
=
another
set of
each circles,
of which
the
touches
#-axis
the
ception ex-
be
at
the
origin. Ex.
2.
Find
the
orthogonaltrajectoryof
the
curves
2
i
A
being the parameter
and
A must
(2)gives
so
that
of the
be eliminated
n\ --
family.
these
between
two
equations.
x(b*+ A)+yyl(a?+ A) 0, =
a2 + A
=
and Thus
the differential
equationof
the
family is
(at-b^yy or
x*-y*+xyyi-
=a2-52 ................
(3)
ORTHOGONAL
Hence
changing y^
TRAJECTORY.
into
differential
the
"
269
of the
equation
,
#1 familv
of
is trajectories 2
But
this
being
primitive,viz.
the
same
the
have
equation (3) must
as
(4) same
:
^
a
n
y2
_-.
i
~
*
i.e.a set of conic sections confocal Find
Ex. 3. cardioides
the
r=a(l"
the former
orthogonal trajectoriesof
0) for
cos
different
set.
the
of
values
family
of
a.
^
Here
r^
and, eliminating a, for the
Hence
with
l~ =
=
dr
0
sm
2
we familyof orthogonaltrajectories
have
must
"
1 dr
n
log
r
or
"
2
log
cos
+
"
constant,
2t
r=b(l+cosO),
or
family of oppositedirection.
another
coaxial
cardioides
whose
point in
cusps
the
EXAMPLES. the
1. Find
of the family of parabolas orthogonaltrajectories
4cM? for different values
#2
=
2.
Show
similar
that
for different values of ellipses-04-^,=m2
3. Find =
a.
orthogonal trajectoriesof
the a2
r
of
the
the
parabolas "
=1
of orthogonal trajectories
the
is s?
of
=Ayb
.
equiangular spirals
a.
of the orthogonal trajectories -f cos
m
family
b2
ae^cota' for different values of 4. Find
the
9 for different values
confocal and of
a.
coaxial
DIFFERENTIAL
270 5. Show
that
EQUATIONS.
the families
of
curves
orthogonal.
are
6. Show r
sin2a
that
the
curves
a(cos 0
=
a)
cos
"
and
r
sinh2/? a(coshft =
"
cos
0)
orthogonal.
are
7. Show
form
that
if f(x+iy)
that
for any cosh
at
+
u
iv
the
curves
orthogonalsystems.
8. Prove
cut
=
x
family /z coth rightangles. the
SOME
220.
The
x
cosec
"
y
"
cosech
of /z the
value
constant
p cot y x
cos
y
=
constant
=
constant
family of
[LONDON, 1890.]
IMPORTANT
DYNAMICAL
EQUATIONS.
equation
general form of the equation of motion particleunder the action of a central force. is the
Multiplying by
2-^and
integratingwe
dO
which
and
we
may
the solution
221.
curves
write
as
is therefore
Equations
effected.
of the form
have
of
a
SPECIAL
SOME
have
alreadybeen
FORMS.
discussed
as
271
beinglinear
with
stant con-
coefficients. The
solution may
Multiplyby
sin
however
n9, which
be conducted
will be found
thus
to be
: "
integrating
an
factor.
Integrating, sin
nO^. d6 -
n(" +
nO=
sin n"dff f*f(ff)
+ A.
Jo
nO is cos Similarly, first integral is cos
cos
nu
factor integrating
an
sin nO=
nu
d\j
and
the
ing correspond-
f'f(ff) nO'dO'+B. cos
J o
Eliminating-^L du
nu
ef(0') sin n(0
=
-
O')d0f+ Bsmn6-A
cos
n6.
0
222. The mass
equationof
often takes
motion
such
some
form
of
a
body
of
changing
as
d! dt ^
and
for this
will equation"f"(x)-rr
be found
to be
factor. integrating
leads at
once
j^x)"l"(x)dxA, i{""(^' J2=
to
+
1
^Xto J
and
the variables
are
separated.
an
DIFFERENTIAL
272
FURTHER 223.
EQUATIONS.
ILLUSTRATIVE
1.
by reducingto alreadydiscussed by
forms
or
Ex.
be solved
Many equationsmay
other of the known artifices. special
one
EXAMPLES.
^=f(ax+by). '
ax
Let
ax+by=z.
Then
=
dx
dx
Thus dx
dz
d*..
x+A=l
or
Ex.2. dx\
y+a
J
+
a
bf(z)
dx
Put
xy=z.
rpi
dy
dz
dx
dx
y+^^=-y-,
Then
.
dz Z
1
, X-j- +-5-
=
dz
dx
or
dx which
is of Clairaut's
form, and
the
completeprimitiveis .
Ex.3.
Solve
e^
-
\dx
dx) Let
Then,
6^ =
since this
equation may dx
77,
he
ex
=
s-
arrangedas \e*dx
ILL
we
write it as
may
which written
USTRA
being of
UTIONS.
SOL
Tl VE
273
?7
Clairaut's form
the
completeprimitivemay
or
Ex.
4.
--
(an. equationoccurringin
Then
and
,v=*Js
Put the
Solid
y
Geometry).
"Jt.
=
equationbecomes
ds
giving
,
t=
ds as
which
is of Clairaut's form
and t-sG
has the
completeprimitive
BC ~
1+2(7'
and
singularsolution
the four
lines straight
9"J-Jy Ex. 5,
Solve the
equation dx
E. I. C.
S
be
DIFFERENTIAL
274 Let
then
the transformation
is known
x
by
EQUATIONS.
be such
direct
that
as integration
a
function
of
t.
dy
d^dL_
Now
dx
*
'
and dx*
f (^ax^yJ^axd4 }dx* dt* dx dt
Thus and
.
the
givenequationthus
reduces
whose
cos
and
of
is
solution is y=A sin qt + B when the value of t in terms
complete. we [Ifa be positive
,
to
"^, x
is
the substituted,
solution
have dx
1
,. "
"
-[=.
""~j= sinh^Wa) If
a
be
negativewe 1
=
t.
have dx
,=dt"
V-a
are
differential Solve the simultaneous Ex. 6. linear with constant coefficients)
equations(which
DIFFERENTIAL
276
EQUATIONS.
whence
operatingupon these in turn by eliminate y and obtain we subtracting, [(D2+ 16)(D2+ 9)+
15
+ 40Z)2 + (Z"4
or
by
3D
and
Z)2" 0, =
1 44"
0,
=
+ 36)^=0, (D*+ 4)(D'2
i.e. whence
x
=
A sin
the Differentiating the second
~2t +
B
2" + C sin 6" + D
cos
whence
we
cos
6*.
three times first equation and subtracting have differential coefficients of y, we
to eliminate
*"
dt
viz.
D2 + 9 and
obtain
the
value
without
of y
any
new
constants,
:
"
y=-%B
sin 2t + 2 A
2t + i"D
cos
sin 6"
-
^-
EXAMPLES. Solve i.
the
equations
2. 2^-(i-*)y"=**. 'cte
3.
4.
5.
(1-
.
2 8. Obtain
the
2
integralsof
cosy
following differential
the
tions equa-
: "
+
9y
-
25
cos
^
[I.C. S, 1804.]
EXAMPLES.
9.
simultaneous
the
Integrate
277
system
4=0. _
10.
Find
of
inclination to
the 11.
as
the
12.
radius
the of
product Find cube
Show of
the of
the
the
curve
current
the
of
the
cosine
in
of
the on
in
which
to
the
tangent
coordinates
form
that curvature
of
form
the
of for
the
inclination
is
7/-axis
of
is
the
y
oc
log
sec ~.
the
proportional
of
constant
the
curvature
the
tangent
projection
length
(l)Soclogtan(?+|), (2)
of
point.
which
for
tangent
#-axis
which
curve
curve
the
the
the
varies to
the
of
the
ANSWEES.
I.
CHAPTER
PAGE
1.
Area
=
e6-ea.
12.
Area=ia2tan
3.
0,
4.
Vol.^. 5
2.
Vol.
Vol.
=-(e*b-e2n).
6.
-a3tan2"9.
=
5.
Vol.=f7ra3.
(a) Vol.
=|
Vol.
=
1 TT
VoL
(8)
*
1 =
-
JL25
Vol.
u
=JL t)
7,
"7Tfia3.
8.
Mass
of
half
the
spheroid
=
J?r/xa262.
279
ANSWERS.
II.
CHAPTER PAGE
17.
a
2.
^Y.
6. 1.
3.
?^1.
7.
x/2-1.
4.
Ioge|.
8.
|. PAGE
"2 X
"'
*'
_c.o
ft C"
r!000
r!00
Ht;
X
+
23.
r!001
"
loo' Tooo' Tool'
_^
_^ 100'
10'
10.
98
PAGE
25.
PAGE
26.
"
2J, 2.
a
logx,
~j
a
log^ +#,
2. 3.
logtan"1^,logsin'1^,log(log^).
(sin6 -sin a).
CALCULUS.
INTEGRAL
280
PAGE
9
4+
"
log2J
aT
log3'
4
+I
log 6
logo2' log tan
4.
_?
""
5.
+
28.
^,
log sin
^
-
cosec
sin-1*,1 tan-,
l86^! an-^,
7. 8.
-log^
+
e*), log(logsin x\
III.
CHAPTER PAGE
32.
1.
sine*, sin#n, sin(log^).
3.
asin^+-tan-1^4,
cose*
-a
+6
log cosh
#.
4 4.
J_ tan-x-JL-. V2
2
_-,
6.
^2
L
.
PAGE /-
2.
cosh^+l),
-1?,
si
sin-V^-
3
"
-
8.
~.
-
"
V*
41. -
g
-H-.-fJain-1*,
'
v
^.
ANSWERS.
3.
-Vl=F, x/^,
281
s
4.
i(^2+1)4,
6.
-^2,| sinh-1^ I siii-1^-2\/r^-iWl
+
2\/l+
#2 +
7.
xyJ\^3?,4 cosh-1-
+
^"?x/5?^
2
8.
^logtan^,
2
-logtana"r+", ^logtan(-+.A 2
CL
\4
/ \
f
13.
),J logtan
x
-
Iog[log{log(loga7)}], log(log^), log{log(log,^)},
CHAPTER
IV. 47.
PAGE
^7
sinh
x
cosh #,
"
x (2+ #2)sinh
x
2"r
sm
"
Zx cosh
x.
2^
CQS
,
in 3a? + 9 sin 3.
J(sin2a?
-
2^
cos
+ cos a?)
3^ + 27
cos^].
2^), sin 4.f
sin 6^7
~""
~""^~~
/cos 2^7 _
cos
4^
cos _
~~
~~
6^7
5.
^sin^
--^sin^-tan-^), 2
6.
CALCULUS.
INTEGRAL
282
v
^2(a2
^2)~sin^-tan-1-,
+
q + r-p,
r+p-q,
sin-1*?+
x/f^;2
p + q-r, 7T2 7T,
of n,
for the values ct /
\
I.
tan-a4).
-
5
-p-q-r.
2
A 7T2-4.
_,
9.
(1 ^2). -
-
339
PAGE
51.
3(^2+ 2)coshx, 5(^4+ 1 2^72+ 24)sinhx. -
(rf+
20^"3+ 1 20#)cosh x
-
6)sinx, _
84\V
\ 2
2/
J{2(2^3 3^)sin2^ -
1.
(a) (m2+ 4
-
+
52.
cos^-sn-g,
where
3
\
3
\
tan
^
log cos
+
(d) ^tan"1^ (a) x
"
"
/
(/)
Jlog(l+^2).
Si
=
^74
1
tan
-"
x
.tT3 3^7 "
,
"
sec"1^
-
_i
lx
"
"
"
cosli"1^.
\/l" ^sin"1^.
+ cos0)-sin"9-logtan (b) 6"(sec(9
(c) 2
where cos (c?) (sin"/" "^" "/"), "
"
\
(e)
x.
A-
/
"
/
(c)^
3)cos
where #=sin#. l)~^rnecos(^-cot-1m),
gsmg
(6)
-
240, 265e-720.
5?r4+ 607T2
PAGE
6^2 +
(2^4
-
-
4
^=
+
Y
where
^
=
sin(9.
CALCULUS.
INTEGRAL
284
CHAPTEK
V.
PAGE
2.
"
.
+ 4# \ log(^'2
3.
+ 2). 5) taii"1^
+
-
4.
-log(3-.r).
5.
#-2log(#2 +
6. 2#
-
2.r +
2)+ 3tan-1(."
+ 6^7 + 10)+ f log(#2
PAGE
^ (iii.)
58.
11
tan-1^ H- 3).
62.
-
_
ct
x (ix.)
"
o
giZ^
7 a^T
+
log(^ q.)+etc. -
-
-
e 1
27
(^Tiy
+
+ 8
(^-l)2 8(^-l)
l0a^"^+ I_L.-?-JL_. g^-l+
16
j B
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