# 9_Integrals.pdf

September 25, 2017 | Author: thinkiit | Category: Integral, Special Functions, Mathematical Objects, Physical Quantities, Numbers

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1

INTEGRALS

INTEGRALS 5.1

BASIC CONCEPTS If f(x) is derivative of function g(x), then g(x) is known as antiderivative or integral of f(x). i.e.,

d dx

ò f(x)dx = g(x).

{g(x)} = f(x) Û

(a)

Derivative of a function is unique but a function can have infinite antiderivatives or integrals.

(b)

ò f( x) dx = g( x) + c, where c is constant of integration, is knwon as indefinite integral. b

(c)

ò f (x ) dx = {g(x) + c }ba = g(b) – g(a).

ò dx = x + c.

(d)

a

(e) (g) 5.2

(f) ò{(x) ± g(x)}dx = òf(x)dx ± òg(x)dx. ò c . f( x)dx = c . ò f ( x) dx . òf(x)dx Û ò f{g(t)}. g’(t)dt, if we substitute x = g(t) such that dx = g’(t)dt.

STANDARD RESULT x n+1 n +1

1.

òx

2.

(ax + b)n+1 ò (ax + b) dx = a(n + 1) + c, n ¹ – 1, n is a rational number..

n

dx =

+ c, n ¹ – 1, n is a rational number. If n = –1, then ò

n

If n = – 1. Then ò 3.

1 dx = log|x| + c. x

1 1 dx = log|ax + b| + c. ax + b a

In case of rational function, if degree of numerator is equal or greater than degree of denominator, then we first divide numerator by denominator and write it as Numerator Denominator

= Quotient +

Re mainder Deno min ator

, and then integrate.

4.

ò sin ax dx =

- cos ax a

5.

ò cos ax dx =

sin ax a

6.

ò tan ax dx = – a log | cos as | + c. or a log | sec ax | + c. If a = 1, then ò tan x dx = – log | cos x | + c or log | sec x | + c.

7.

ò cot ax dx = a log | sin ax | + c.

8.

ò sec ax dx = a log | sec ax + tan ax | + c.

9.

ò cosec ax dx = a log | cosec ax – cot ax | + c. If a = 1, thenò cosec x dx = log | cosec x – cot x |+c.

10.

ò sec ax tan ax dx = a sec ax + c.

11.

ò cosec ax cot ax dx = – a cosec ax + c.

12.

ò sec

13.

ò cosec

14.

òe

+ c. If a = 1, ò sinx dx = – cosx + c. If a = 1, ò cos x dx = sin x + c.

+ c.

1

1

1

If a = 1, then ò cot x dx = log | sin x | + c.

1

If a = 1, then ò sec x dx = log | sec x + tan x | + c.

1

1

1

2

ax dx = 2

ax

1 tan ax + c. a

ax dx = –

dx =

eax a

If a = 1, then ò sec x tan x dx = sec x + c. If a = 1, then ò cosec x cot x dx = –cosec x + c.

If a = 1, then ò sec2 x dx = tan x + c.

1 cot ax + c. If a = 1, then a

ò cosec

2

x dx = –cot x + c.

+ c. If a = 1, then ò ex dx = ex + c.

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2

5.3 (i)

INTEGRALS amx

ax

15.

ò amx dx = m loge a + c. If m = 1, then ò ax dx = loge a + c.

16.

ò

17.

òa

18.

òx

19.

ò

21.

òx

2

23.

ò

a2 - x2

dx =

x 2

a2 - x2

+

a2 2

sin–1

24.

ò

a2 + x2

dx =

x 2

a2 + x2

+

a2 2

2 2 log x + a + x + c.

25.

ò

x 2 - a2

dx =

x 2

x 2 - a2

a2 2

1 a -x 2

2

1 + x2

dx =

1 x -a 2

1 2

a +x

2

1 - a2

x a

dx = sin–1

2

1 x tan–1 a a

1 - x2

+ c. If a = 1, then ò

1 x sec–1 a a

dx =

2

1

+ c. If a = 1, then ò

1 1+ x2

+ c. If a = 1, then

2 2 dx = log x + a + x + c.

x-a

1

dx = 2a log x + a + c. x a

òx

dx = sin–1x + c. dx = tan–1x + c. 1 x2 - 1

dx = sec–1x + c. 1

20.

ò

22.

òa

x - a2 2

2

1 - x2

x 2 - a2

dx= log x +

+ c.

a+x

1

dx = 2a log a - x + c.

+ c.

log x +

x 2 - a2

+ c.

METHODS OF INTEGRATION Integration by Substitution (or change of independent variable) : If the independent variable x in ò f(x) dx be changed to t, then we substitute x = f (t)

ò f(x) dx = ò f(f(t))f’(t) dt

i.e., dx = f‘(t) dt (ii)

which is either a standard form or is easier to integrate. Integration by parts : If u and v are the differentiable functions of x then (a) (b) (c) (d)

ò u. v dx = u ò vdx – ò êëçè dx u ÷ø(ò vdx )úû dx. éæ d

ù

ö

How to choose Ist and IInd functions: If the two functions are of different types take that function as Ist which comes first in the word ILATE where I stands for inverse circular function, L stands for logarithrmic function A stands for Algebratic function, T stands for trigonometric function and E stands for exponential function. If both functions are algebratic take that function as Ist whose differential coefficient is simple. It both functions are trigonometrical take that function as IInd whose integral is simpler. Successive integration by parts : Use the following formula

ò uvdx n= uv1 – u’v2 + u”v3 – u’” v4 + ......... .........+ (–1)

n–1

u

n –1

n

vn + (–1)

òu

n

vndx.

where u stands for nth differential coefficient of u w. r. t. x and vn stands for n th integral of v w. r. t. x. cancellation of Integrals : i.e. ò e {(f(x) + f’(x)} dx = e f(x) + c. x

(iii)

x

Evalution of Integrals of various Types : Type I : Integrals of the form

(i) ò

dx ax + bx + c 2

(ii) ò

(ax

dx 2

+ bx + c

)

(iii) ò (ax 2 + bx + c ) dx

2

Rule : Express ax + bx + c in the form of perfect square and then apply the standard resuts. Type II : Integrals of the form

(i) ò

px + q 2 ax + bx + c

(ii) ò

px + q ax + bx + c 2

(iii) ò

(p

0

xn + p1xn-1 + ...... + pn-1x + pn ax 2 + bx + c

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(

)

) dx

3

INTEGRALS Rule for (I) : ò

(px + q)dx ax 2 + bx + c

=

p 2a

(2ax + b )dx

b

æ

pb ö

dx

= (2a ) ò (ax 2 + bx + c ) + ç q - 2a ÷ ò 2 è ø ax + bx + c æ

bp ö

è

ø

dx ax 2 + bx + c

2 ln | ax + bx + c | + ç q - 2a ÷ ò

The other integral of R. H. S. can be evaluated with the help of type I.

( px + q) dx

Rule for (II) : ò

ax2 + bx + c

=

(ax

p a

2

p 2a

=

ò

( 2ax + b ) dx

(ax

2

+ bx + c

bp ö æ + bx + c + ç q - 2a ÷ è ø

)

)

æ

pb ö

è

ø

dx

dx + ç q - 2a ÷ ò

ax + bx + c 2

dx

ò

ax + bx + c 2

The other integral of R. H. S. can be evaluated with the help of type I. Rule for (III) : In this case by actual division reduce the fraction to the form f(x) +

(px + q)dx ax 2 + bx + c

and then integrate. dx

dx

dx

(i) ò a + b sin 2 x

Type III : Integrals of the form

(ii) ò a + b cos2 x

(iii) ò asin2 x + bcos2 x

f(tan x )dx

dx

(iv) ò (a sin x + b cos x )2

(v) ò a sin 2 x + b sin x cos x + c cos2 x + d

where f (tan x) is a polynomial in tan x. 2 Rule : We shall always in such cases divide above and below by cos x ; then put tan x = t i.e. sec x dx = dt then the question shall reduce to the forms ò 2

(at

dt 2

+ bt + c

)

or ò

dx

)

.

(ii) ò a + b cos x

(p cos x + q sin x )

dx

(iii) ò a sin x + b cos x + c

+ bt + c

dx

(i) ò a + b sin x

Type IV : Integrals of the form

(at

j ( t ) dt 2

p cos x + q sin x + r

(iv) ò (a cos x + b sin x ) dx (v) ò a cos x + b sin x + c

Rule for (i), (ii), and (iii): write

cos x =

1 - tan2 x / 2 , 1 + tan2 x / 2

sinx =

2 tan x / 2 1 + tan2 x / 2

2

The numberator shall become sec x/2 and the 2

denominator will be a quadratic in tan x/2. Putting tan x/2 = t i.e. sec x/2 dx = 2dt the question shall reduce to the form r

ò at

2

dt + bt + c

. Rule for (iv) : express the numberator as

r

I(D ) + m(d.c. of D ) find l and m by comparing the coefficients of sin x and cos x and split the integral into sum of two integrals as l ò dx + m

ò

d.c. of Dr Dr

r

= lx + m ln | D | + c

Rule for (v) : Express the numberator as r r l(D ) + m (d.c. of D ) + n, find l, m, and n by comparing the coefficients of sin x, cos x and constant term and split the integral into sum of three integrals as l

ò

dx + m

ò

d.c. of Dr Dr

dx + n ò

dx Dr

r

= lx + m ln | D | + n

dx

òD

r

and to evaluate

method to evaluate rule (i), (ii) and (iii).

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dx

òD

r

proceed by the

4

INTEGRALS Type V : Integrals of the form

(x

)

+ a2 dx x 4 + kx 2 + a 4

ò(

(i)

2

)

(x

)

- a 2 dx x 4 + kx 2 + a 4

ò(

(ii)

where k is a constant, + ve, –ve or zero.

2

2

Rule for (i) and (ii) : Divide above and below by x then putting (i) t = x – i.e.,

dt =

æ a2 ö ç1 + 2 ÷ dx and ç x ÷ø è

dt =

)

a2 x

æ a2 ö ç1 - 2 ÷ dx ç x ÷ø è

then the questions shall reduce to the form

òt

2

dt + c2

or

òt

2

dt - c2

Remember : (i)

x 2dx x 4 + kx 2 + a 4

ò

(ii) ò

(x

=

dx 4

2

+ kx + a

4

1 2

)

=

(x

)

+ a2 dx x 4 + kx 2 + x 4

ò(

1 2a2

dx

2

(x

)

+

)

1 2

+ a2 dx x + kx 2 + x 4

ò(

2

4

2

1 2a2

(2n - 3 )

x

)

- a 2 dx x 4 + kx 2 + a 4

ò(

)

(x

(x

) )

- a 2 dx x + kx 2 + a 4

ò(

2

4

)

dx

(iii) ò (x 2 + k )n = k (2n - 2)(x 2 + k )n-1 + k (2n - 2) ò (x 2 + k )n-1 Type VI : Integrals of the form

(i) ò

dx

(ii)

(Ax + B) (ax + b )

(iv) ò

( Ax

(

dx 2

Ax + Bx + C

dx

2

+ Bx + c

) (ax

2

+ bx + c

(iii) ò

dx

( Ax + B )

( ax

2

+ bx + c

)

)

Rule for (i) and (ii) : Put ax + b = t

(iv)

)

( ax + b )

Rule for (iii) :

Put Ax + B =

Rule for (iv) :

Put

2

1 t

ax2 + bx + c Ax2 + Bx + C

Integration by partial fraction : Form of the Rational Function

2

=t

Form of the Partial Fraction

1.

px + q (x - a)(x - b ) ,

2.

px + q (x - a)2

3.

px 2 + qx + r (x - a )(x - b )(x - c ) ;

A B C + + x -a x -b x -c

4.

px 2 + qx + r (x - a )2 (x - b)

A B C + + 2 x - a (x - a ) x-b

5.

px 2 + qx + r (x - a) x 2 + bx + c

a¹b

A B + x -a x -b

A B + x - a (x - a )2

;

(

);

A Bx + C + x - a x 2 + bx + c

2

where x + bx + c cannot be factorised further. In the above table, A, B and C are real numbers to be determined suitably.

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5 5.4

INTEGRALS PROPERTIES OF DEFINITE INTEGRALS (i)

a

ò a f ( x ) dx = 0

(ii)

b

c

b

(iii)

ò a f ( x ) dx = ò a f ( x ) dx + ò c f ( x ) dx (a < c < b)

(v)

ò 0 f ( x ) dx = ò 0 f (a - x ) dx

a

(iv)

b

a

ò a f ( x ) dx = – òb f ( x ) dx .

b

b

ò a f ( x ) dx = ò a f (a + b - x ) dx

a

(vi)

2a

ò0

f ( x ) dx

=

a

ò 0 f ( x ) dx

a

ò 0 f ( 2a - x ) dx

5.5

2a

a

2a

(vii)

ò0

(viii)

ò-a f ( x) dx = 2 ò0 f ( x ) dx , if f(x) is even function, i.e., f(–x) = –f(x).

f ( x ) dx

= 2 ò 0 f ( x ) dx ,if f(2a-x) = f(x) and

a

ò0

f ( x ) dx

= 0, if f(2a-x) = – f(x)

a

INTEGRATION AS A LIMIT OF SUMS. b

ò a f ( x ) dx = or

Lt

h®0

h[f(a) + f(a + h) + f(a + 2h) + ................+ f(a + n - 1 h)]

= (b – a) Lt

h® ¥

1 n

[f(a) + f(a + h) + f(a + 2h) + ..........+ f(a + n - 1 h)], where h =

The following results are used for evalating questions based on limit of sums. (i) 1 + 2 + 3 + ................+ (n – 1) = å(n–1) = (ii) 1 + 2 + 3 + ..........+ (n–1) = å(n–1) = 2

2

2

2

2

( n - 1) n 2 (n - 1)n(2n - 1) 6

(iii) 1 + 2 + 3 + .........+ (n – 1) = å(n–1) = 3

3

3

3

n–1

(iv) a + ar + ..........+ ar

=

[

3

é (n - 1)n ù ê 2 ú ë û

2

] (r ¹ 1).

a rn - 1 r -1

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b-a n

+

6

INTEGRALS

SOLVED PROBLESM Ex.1

Find f(x), if 3 (i) f’(x) = (x – 1) and f(2) + f(–2) = 0

(ii) f’(x) = a (cos x + sin x ), f(0) = 9 and

æp ö

f ç 2 ÷ = 15 è

Sol.

ø

Here, f ’(x)

\

f(x)

= ò (x–1) dx = (x – 1) + C

\

f(2)

= (2 – 1) + C = =

Þ

1 4

= (x – 1)

3

(i)

1 4

3

1 4

81 + 4

+C+

4

4

Þ

C=0

1 4

82 4

+C

Given that f(2) + f(–2) = 0, we have

+ 2C = 0

Þ 20.5 + 2C = 0

Hence, from (1) we have 1 4 (x – 1) – 10.25 4

f(x) = (ii)

Here, f ’(x)

\

f(x)

=

1 4 [(x – 1) – 41] 4

= a (cos x + sin x)

= ò a (cos x + sin x)dx

= a ò cos x dx + a ò sin x dx = a sin x – a cos x + C ...(1) \ f(0) = 9 = a×0 – a×1+C ...(2) Þ C–a=9 æ pö

f ç 2 ÷ =15 = a×1 – a × 0 + C Þ a+C =15

and

è ø

From (2) and (3), we have a = 3, C = 12 Hence, from (1), we have f(x)=3(sin x–cos x)+12 Ex.2

Integrate the following functions w.r.t. x : 2 1 + log x ) ( x + 1 )( x + log x ) (i) ( (ii)

2

x

x

(iii) x x + 2 Sol.

(i) Put 1 + log x = y. Then, So, I = Þ

òy

2

1 dx = dy x

(1 + log x )2 dx ò x

dy =

y3 3

+C=

(1 + log x )3 + C 3

æ

è

ø

(ii) Put x + log x = y. Then, ç1 + x ÷ dx = dy So,

I=

ò

æ x + 1ö

Þ ç x ÷ dx = dy è ø

(x + 1)(x + log x )2 dx x

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Þ C = – 10.25

7

INTEGRALS y3 3

= ò y dy = 2

(iii)

(x + log x )3 + C

+C=

3

So, I = ò x x + 2 dy = ò (y - 2)y 2 dy 1

Put x + 2 = y. Then, dx = dy

Þ ò ( y 2 - 2y 2 ) dy = ò ( y 2 dy - 2)y 2 dy 3

Þ

3

1

5

2

2 2 2 y - 2. y 3 + C 3 5

Ex.3

Find : (i)

Sol.

(i)

5 3 2 (x + 2)2 - 4 (x + 2)2 + C 3 5

=

ò cos x dx 3

1

ò

(ii)

1 - sin x dx

3

We have

cos x =

1 4

(cos 3x + 3 cos x) \

ò cos

3

x dx =

1 4

cos xdx =

1 sin 3x 4 3

(ii)

3 4

+ sinx + C =

x x xö æ 2 x + sin2 ÷ - 2 sin cos ç cos 2 2 2ø 2 è

xö x æ ç cos - sin ÷ 2ø 2 è

=

ò

\ Find : (i) (ii) Sol.

(i) \ (ii) \

3 4

sin3x + sinx + C

We have 1- sin x =

Ex.4

1 12

ò

2

x x = cos - sin 2

x2

ò

x 6 + a6

sec 2 x

x 2

dx –

x

ò sin 2 dx

= 2 sin

x 2

+ 2 cos

3

x -1

ò

+C

dx

x2 - 1

2

1 3

2

Put x = t so that 3x dx=dt or x dx = dt

ò

x2 x 6 - a6

dx =

1 3

dt

ò

( )

t 2 + a3

2

1 1 = log t + t 2 + a 6 + C = log x 3 + x 6 + a 6 + C 3

3

2

Put tan x = t so that sec x dx = dt

ò

sec 2 x tan x + 4 2

dx =

dt

ò

t2 + 4

2 = log t + t + 4 +C

= log tan x + tan2 x + 4 +C (iii)

x 2

dx

dx (iii)

tan2 x + 4

2

ò cos

1- sin x dx =

ò

x -1 x -1 2

dx=

1 2

ò

2x x -1 2

dx – ò

dx x2 - 1

dx

2 2 = x - 1 - log x + x + 1 + C

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3

ò cos 3x dx + 4 ò

8

INTEGRALS x +2

Ex.5

Find : (i) ò

Sol.

(i) Let x + 2 º A.

x + 2x + 3

d dx

5x + 3

dx (ii) ò

2

dx

2

x + 4 x + 10

2

(x + 2x + 3) + B

Þ

x + 2 º A.(2x + 2 ) + B

Þ

5x + 3 º A (2x + 4) + B

On comparing coefficients of like powers of x, we have 1 2

1 = 2A and 2 = 2A + B Þ A = So, ò

x+2 x 2 + 2x + 3 2x + 2

1 2

=

1 ×2 x 2 + 2x + 3 + C1 2

ò

x + 2x + 3 2

1 (2x + 2) + 1 2 dx x 2 + 2x + 3

dx = ò

=

and B = 1

dx

dx + ò

x + 2x + 3 2

+

dx

ò

(x + 1)2 + (

2

)

2

= x 2 + 2x + 3 + C1+log x + 1 + x 2 + 2x + 3 +C2 = x 2 + 2x + 3 + log x + 1 + x 2 + 2x + 3 + C (ii) Let 5x + 3 º A.

d dx

2

(x + 4x + 10) + B

On comparing coefficients of x, we have 5 = 2Aand 3 = 4A + B Þ A =

ò

\

ò

5x + 3 x 2 + 4x + 10

and B = – 7

dx

5 (2x + 4) - 7 5 2 dx = 2 x 2 + 4 x + 10 =

5 2

ò

2x + 4 2

x + 4x + 10

5 ´ 2 x 2 + 4 x + 10 + C1 - 7 2

dx - 7

ò

dx

ò

( x + 2) 2 +

dx 2

x + 4 x + 10

( 6 )2

= 5 x2 + 4x +10+C1 -7log (x + 2) + x2 + 4x +10 +C2 = 5 x 2 + 4 x + 10 - 7 log x + 2 +

x 2 + 4 x + 10 + C

dx

ò ( x + 1)( x + 2 )

Ex.6

Find :

Sol.

We write

1 ( x + 1)( x + 2)

º

A x +1

B x+2

where real numbers Aand B are to be determined suitabley. This gives 1 º A (x + 2) + B(x + 1) Equation the coefficient of x and the constant terms, we get A + B = 0 and 2A + B = 1 Þ A = 1 and B = –1 Thus,

1 ( x + 1)( x + 2)

=

1 x +1

+

-1 x+2

= log | x+1 | – log | x+2 | + C = log

x -1 x+2

Þ

dx

ò ( x + 1)( x + 2) = ò x 1+ 1 dx - ò x 1+ 2 dx

+C

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9

INTEGRALS

Ex.7

Find : (i)

Sol.

(i)

3x - 1

x

ò ( x + 2 )2 dx

Let 3x - 12 º ( x + 2)

(ii) ò (x - 1)2 )(x + 2) dx

A x+2

B

+

( x + 2)2

Then, 3x – 1 º A(x + 2) + B Comparing coefficients of x and the constant terms on both sides, we get 3 = A and –1 = 2A + B Þ A = 3 and B = – 7 3x - 1

1

7

3

Thus,

ò ( x + 2)2 dx = ò x + 2 dx - ò ( x + 2)2 dx = 3 log |x + 2| – 7 . ( x + 2)

(ii)

Let

Þ

A=

Thus,

ò (x - 1)2 ( x + 2) dx

x

º

A x -1

+

B

+C

C

+ ( x - 1) x+2 ( x - 1) ( x + 2) 2 Then x =A(x – 1) (x +2) + B(x + 2) + C (x + 1) x = A (x2 + x – 2) + B (x + 2) + C (x2 – 2x + 1) Comparing coefficients of x2, x and the constant terms on both sides, we get 0 = A + C; 1 = A + B – 2C ; 0 = –2A + 2B + C

Ex.8

2

2 9

;B =

1 ;C 3

=–

2

2 9

x

2 1 2 9 dx + 3 9 dx dx x -1 x+2 ( x - 1)2

=

ò

ò

ò

=

2 9

log |x – 1| –

=

2 9

log x + 2 – 3( x - 1) + C

1 3( x - 1)

x -1

2 9

log |x + 2| + C

1

Find : (i) ò x sin-1 x dx (ii) ò x cos-1 x dx

Sol.

(i) = =

Put x = sin t so that d x = cos t . dt

1 2

So,

ò x sin

-1

ò

x dx = sin t . t . cos t dt

ò t . sin 2t dt

ò[ ò

] ]

1 t . sin 2t dt - 1. sin 2t dt dt 2

= 1 éê t . cos 2t + ò cos 2t dt ùú 2ë

-2

2

û

=– 1 t . cos 2t + 1 . sin 2t + C 4

=

4

2

1 1 t .(1 - 2 sin 2 t ) + sin t 1 - sin 2 t + C 4 4

= (ii)

1 1 (2x 2 - 1) sin -1 x + x 1 - x 2 + C 4 4

Put x = cos t so that dx = – sin t dt

So,

ò x cos

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-1

ò

x dx = - cos t . t . sin t dt

10

INTEGRALS = - 1 ò t sin 2t dt = 1 t . cos 2t - 1 . sin 2t + C 2

= Ex.9

1 1 t(2 cos2 t – 1) – 4 4

1 - cos 2 t

(Using (1) give below)

2

4

4

1 x (2x2 – 1) cos–1 x – 4 4

. cos t + C =

1- x 2

+C

é 1 1 ù ú dt 2 ëê logx (logx) úû

Find : (i) ò ê

æ sin4x - 4 ö

(ii) ò e x ç ÷ dx è 1 - cos4x ø Sol.

(i)

Let log x = t. Then x = et and

= ò e t [f ( t ) + f ' (t ) dt, where f ( t ) = = et .

1 t

= dt \

é 1

1

ù

= et . f(t)+C

x

x æ sin 4 x - 4 ö

ç ÷ dx è 1 - cos 4 x ø

4 é sin 4x ù ú 1 cos 4x 1 cos 4x ë û

= ò ex ê

é 2 sin 2x cos 2x

= ò ex ê

2

ë

2 sin 2x

[

dx

ù ú dx 2 sin 2x û 4

-

2

]

= ò e x cot 2x - 2 cos ec 2 2x dx = ò e x [f(x) + f '(x) dx , where f(x) = cot 2x = ex . f(x) + C = ex . cot 2x + C Ex.10 Find : (i) (ii) Sol.

(i)

ò

ò

x 2 - 8x + 7 dx

1 - 4x - x 2 dx

I=

ò

x 2 - 8 x + 7 dx

ò

( x - 4 )2 - 3 2 dx

I=

ò

1 - 4 x - x 2 dx =

=

ò

- é(x + 2)2 - 5 ù dx ë û

=

=

2 ò [( x - 4) - 9] dx

x-4 2

=

x 2 - 8x + 7

9 2

log ( x - 4) + x 2 - 8 x + 7 + C

(ii)

1

ò

- ( x 2 + 4 x - 1) dx

= 5

ò

æ1

ò êêë log x - (log x)2 úúû dx = ò ççè t - t2 ÷÷ø e

+ C = log x + C

òe

(ii)

1 t

dx x

5 - ( x + 2)2 dx

æ x +2ö ÷÷ + C 5 ø

= (x + 2) 1 - 4 x - x 2 + sin–1 çç 2 2 è

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t

dx

11

INTEGRALS

Ex.11 Evaluate te following definite integrals as limit of sums : Sol.

b

where

lim

b-a n

h=

Here, a = 0, b = 2, f(x) = x and 2-0 n

h= \

ò0 x dx

òa f (x ) = h ® 0 h{f(a) + f(a + h) + f(a + 2h) +........+ f[a + (n – 1)h]}

We have

(i)

2

2

2 n

=

or nh = 2

lim

ò0 x dx = h ® 0 h{f(0) + f(0 + h) + f(0 + 2h) + .......+ f[0 + (n – 1)h]}

= hlim ® 0 h{0 + (0+h) + (0+2h) +....+ [0 + (n–1)h]} = hlim ® 0 h[h + 2h +......+ (n – 1)h] 2

æ 2 ö é (n - 1) n ù 2 úû è ø

lim ç ÷ 2 = hlim ê ® 0 h [1+2+....+ (n – 1)] = h ®¥ n ë

2 n

(Q h =

and as h ® 0, n ® ¥)

é æ 1 öù = nlim ® ¥ ê2ç1 - ÷ ú = 2 n øû

ë è

π /2

ò0

Ex.12 Show that π/4

ò0

= 2 Sol.

p 4

Let x =

f(sin2x) sin x dx

f(cos2x)cosx dx. p 4

– t. Then t =

When x = 0, t =

p 4

–0=

p 4

– x Þ dt = –dx

òp / 4

=

ò-p/ 4 f(cos 2t) çè sin 4 cos t - cos 4 sin t ÷ø dt

=

ò-p/4 f(cos 2t) çè

p/ 4

æ

p/ 4

æ 1

1 2

.2

p 2

=–

p 4

\

p /4

ò0

p

2

p

f (sin 2x ) sin x dx

cos t -

ö

ö sin t ÷ dt 2 ø

1

f (cos2t)cos t.dt -

1 2

=

1

p /4

f(cos 2t)cos t dt 2 ò-p /4

1

p/4

f(cos 2t)sin t dt 2 ò-p /4

.0

[Q f (cos 2t) cos t is an even function ; and f (cos 2t) sin t is an odd function] =- 2

p/2

ò0

ì æ p öü æp ö f ísin 2ç - t ÷ý sin ç - t ÷ ( -dt ) 4 è øþ è4 ø î

-p / 4

=

=

p p ,t= 2 4

and when x =

p /4

ò0

f (cos 2t)cos t dt

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12

INTEGRALS

UNSOLVED PROBLEMS EXERCISE – I cos ec x

Q.1.

Evalute :

(i) ò cos ec x - cot x dx

Q.2

Evalute :

-1 (i) ò tan ç 1 + cos 2x ÷ dx

Q.3

(i) If f¢(x) = sin2x + cos3x + 5, f(0) =

æ

sin 2x ö

è

ø

(ii)

3 2

ò

sin 6 x + cos 6 x dx sin 2 x. cos 2 x

(iii) ò 1 + cos 3 x dx

(ii) ò tan -1(sec x + tan x) dx , –

p p