Jee 2014 Booklet5 Hwt Solutions Integral Calculus 1
Short Description
Jee 2014 Booklet5 Hwt Solutions Integral Calculus 1...
Description
Vidyamandir Classes
Solutions to Home Practice Test/Mathematics Integral Calculus-1 1.(C)
I
n cos n 1
1 sin x
n
n sec x sec x tan x dx
sec x tan x n sec xdx
dx
HWT - 1
sec x tan x n 1
n
Put sec x tan x t and sec x sec x tan x dx dt
2.(CD)
I
t
ndt
n 1
1 t
n
1
sec x tan x
cos n x
n
1 sin x n
tan 2 x . tan 3x . tan 5x dx tan 5x tan 2 x tan 3x dx
[Using tan 5 x tan 2 x 3x ]
1 1 1 log sec 5 x log sec 2 x log sec 3x k 5 2 3
log sec 1 2 2 x . sec 1 3 3 x . sec1 5 5 x k a
3.(D)
x x x Put 2 x I
3t
5.(D)
x 2t x t
2x
3x t 6
2t
2 x
1t
x 2t 1 x t 1
3t
1 5
and
a 3 b 3 c 3 3a 1b 1c 1
1t
dx
1t
3x 2t 6 x t
ab bc ca 0
dx
3t 1
dx
3x 2t 6 x t u and 6t x 3t 1 x 2t 1 x t 1 dx du I
u 1 t du u 1 t 1
1 6t 1 t
6t
2x
3t
3x 2t 6 x t 6 1 t
Put xe x t, e x xe x dx dt to get :
cos xe cos 2
I
dt
x
x 1 dx
x 1 xe x
I
2
2
t
sec
dt
2
e x dx
x
t 1 t 1 t dt
dt
2
1 t t dt 1 t 1 t dt
1
dt
dt
2
2
1 1 t c log c 1 t 1 t 1 t
1 c 1 xe x
2 e e 1 . Put e x
t dt tan t tan xe x c
1 t t dt 2 t 1 t
1 t t dt t 1 t
xe x log 1 xe x
I
2
. Put xe x t to get e x x 1 dx dt
t 1 t
log t log 1 t
7.(B)
x 2t 1 xt 1 x 2 x 2t 3 xt 6
e x 1 x dx
6.(B)
c
and
3t 1
3t
1 1 ,b 2 3
x
t to get: e x dx dt
t 2 t 1 dt dt t 2 t 1 t 2 t 1
I
1
1
t 1 t 2 dt
ex 1 t 1 n t 1 n t 2 c n c n x c e 2 t2
VMC/Solutions/Integral Calculus-1
43
HWT-5/Mathematics
Vidyamandir Classes 8.(B)
I
x 5
1
9.(D)
x4
I
e
x
dx . Put x 4 t 2 to get dx 2t dt
t 1 t 2 t 2t dt
2
dt 2
2 tan 1 t c 2 tan 1
1
x4 c
sec x e x . log sec x tan x dx e x log sec x tan x sec x dx d log sec x tan x dx e x . log sec x tan x c e x log sec x tan x dx
e x . log sec x tan x c 10.(A)
I
sin 1 x cos 1 x 1 dx sin x cos 1 x
4 sin 1 x 1 dx .
4t
sin 2
1
x cos 1 x dx
2
2 sin
1
x
Put x sin 2 t to get dx sin 2t dt
1 sin 2t dt t sin 2t dt sin 2t dt
I
4
4 t cos 2t cos 2t cos 2t 2 sin 2t 4 cos 2t dt t cos 2t c 2 2 4 2 2
2 1 2 x 1 x 4 1 2 x 2 sin x 1 2 x c sin 1 x 2 x 1 x 1 x x c 4 2
Integral Calculus-1 5.(B)
x log 1
ex
dx 2
e log
e2 x
e dx
1
HWT - 2
x log e log x 2 log e log x dx x 1 log x 2 log x dx log e
log e
1
1
1
Let log x k
e
1
k
1 1 ek dk 1 k 2 k
1
1
1 k 2 k dk
1 k log 2 k f x k log x
lim
x0
6.(D)
lim
x0
log x 1 0 0 form from L.H. rule x
1 1 x 1
Let x 3 t 2
t 1 t t
7.(CD)
1 t 1 2 log t 1 2 1
2t dt
2 dt
2
2
log
x 3 1 x 3 1
Let 1 x3 t 2
xdx 1 x3
2dt
3x
2 cos 1 1 x3 3
32
2 3
dt 1 t2
2 cos 1 t 3
But cos 1 1 x3 can also be written as sin 1 x3 2
2
2 x log 2 dx 2 x e x log 2 I
I
9.(B)
Form perfect square in denominator
10.(B)
Let e x k
x
e x dx 2 x e x
e
8.(B)
x
e
x
f x sin 1 x and g x x3 2 x x
I
2x . ex c 1 log 2
sin e x dx sin k dk cos k cos e x
VMC/Solutions/Integral Calculus-1
44
HWT-5/Mathematics
Vidyamandir Classes
Integral Calculus-1 1.(C)
e
x
tan x log cos x e x log cos x c e x log sec x c
3.(C)
e x has range R+.
Let x 2 t
Let
6.(D)
x
8.(D)
1 2
1 4
x 1
34
4
t
1
1
2
x 1 2
2
34
1 x5 1 x4
34
dx
1 t1 4 C t1 4 C , where t 4 14
dt
14
1 1
1
1 1 x4
C
x4
x sin x
1 cos x dx x 2 sec Let I
2x
1 4
x
2
x x tan dx 2 2
dx
1 log 2
x 2 sec 1
1 1 t2
2
K
dt
1 1 t sin 1 C sin 1 2 x C log 2 1 log 2
1 . log 2
Putting tan 1 x t and
etan
1 x
dx 1 x2
1 x x2 1 x2
dt , we get : 1 dx et tan t sec 2 t dt et tan t C etan x . x C
g x f x f x dx g x f x dx g x f x dx f x g x dx f x g x dx dx g x f x dx f x g x g x f x dx g x f x dx C g x dx g x f x g x C Integral Calculus-1
1.(D)
x x x x x x dx tan dx x tan tan dx tan dx C x tan C 2 2 2 2 2 2
2 x t, 2 x log 2dx dt I
10.(C)
dx
ex
x t
Putting
9.(A)
x e
5.(A)
dx
t t tet et e dt 2 2
3 x2
7.(A)
HWT - 3
HWT - 4
sin x 4 sin x 1 dx 1
1 5
sin x 4 sin x 1 sin x 4 sin x 1
1 5
2t 4 1 t 2 dt 1 dt 1 5 t 2t 1 10 t t 1
1 5
2t 1 t
dx
2dt
2
1 5
2
2dt
2
sin x 1 dx 5 sin x 4 dx 1
1
1
x Putting tan 2 t
2
2
VMC/Solutions/Integral Calculus-1
45
HWT-5/Mathematics
Vidyamandir Classes 2 5
t 1 1
1 10
dt
2
1
15 1 t 4 4 2
2
dt
4t 1 2 1 2 tan 1 C 5 t 1 5 15 15
x 4 tan 2 1 1 2 1 tan C x 15 tan 1 5 15 2 x 4 tan 1 2 2 2 A ,B , f x 5 5 15 15
2 . 5
2.(C)
1 2 tan x tan x sec x dx 1 tan x tan x 2 tan x sec x dx sec x tan x 2 tan x sec x dx tan x sec x dx log sec x log sec x tan x C log sec x sec x tan x C tan x tan x I sin x cos x dx tan x sec x dx 1 t dt , where t tan x 12
2
3.(A)
12
2
2
12
2
2
I 2t1 2 C 2 tan x C
4.(C)
Putting
5x
t , we have
55
x 55 5 x x
5 5 log 5 dx dt , we get :
5
5
55
5.(A)
x
3
5x
.5
. 5 dx x
1
log 53
3
C
55
log 53
C
1 sin x dx 1 cos x dx 1
1
2
1 x x sec 2 dx tan b 2 4 2 4 2
x tan b , 4 2 a
6.(C)
1 dt log 5
5x
t
where b = const.
,bR 4
f x g x f x g x dx
f x g x dx f x g x dx f x g x f x g x dx g x f x g x f x dx
f x g x f x g x
7.(C)
4e x 6e x
9e
x
4e
x
dx
4e2 x 6
9e
4e2 x
2x
4
dx
9e
2 6 log 9e2 x 4 log 9 4e 2 x C 9 8
2 3 3 log 9e 2 x 4 log 9e 2 x 4 log e 2 x C 9 4 4
3 35 x log 9e 2 x 4 C 2 36
2x
4
dx 6
9e
18e2 x
1 4
2x
VMC/Solutions/Integral Calculus-1
dx
2 9
9e
2x
4
dx 6
e2 x
9 4e
2 x
dx
46
HWT-5/Mathematics
Vidyamandir Classes 8.(B)
x
I
Putting
x 1 t 2 , dx 2t dt , we get :
I 2
x2 1
x
4
1
t
2
9.(B)
x2
Let
3
2
3x 3
t 1 2
4
tan
dx
x 1
dt 2
t2 1
1
1 1 t
2
dt
2
1 t t 3
1 t t 2 x tan 1 C 3 3 3 x 1
1 1 x x 1 x dx 1 1 x dt 1 dx t 2
C
2
dx
2
2
2
2
where t x
2
x 2 2 x
x2 1 t 1 tan 1 tan 1 C C 2x 2 2 2
1
10.(A)
Putting,
1 x
l r 1 x t and
xl
1 xl x l
1 dx 2
x . . . .l r x
2
x l 3 x . . . .l r x
dx dt , we get :
1 . dt t C l r 1 x C
Integral Calculus-1
HWT - 5
1.(B) 2.(A)
Putting x n 1 t and n x n 1dx dt , we get : 1 1
1
x x 1 dx n t t 1 dt n t 1 t dt 1
1
1
n
3.(B)
x t and
Putting
4.(BC)
a
x
x
xn 1 1 t 1 log C log C xn 1 n n t
1 2 x
dx dt , we get :
dx 2 a t dt
2a t 2a x C C log a log a
5 4 cos x dx 1
Hence,
x 2 dx 5 1 tan 2 x 2 4 1 tan 2 x 2
1 tan 2
t
2dt 2
9
, where t tan
x 2
2 2 tan x 2 t tan 1 C tan 1 C 3 3 3 3
A
2 1 and B 3 3
VMC/Solutions/Integral Calculus-1
47
HWT-5/Mathematics
Vidyamandir Classes
5.(AC)
log x 1 log x x x 1
dx
1 log 1 x dx 2 x x
1 log 1 x log t dx dt , t 1 x 2 1 x
where t 1
1 x
2
6.(A)
2 1 1 1 1 log t 2 C log 1 C log x 1 log x C 2 2 x 2
1 log x 1 2
x tan 1 x 1 x
2
2 2 log x 2 2 log x 1 . log x C 12 log x 1 12 log x 2 log x 1 . log x C
1 2
dx
tan
1
x.
2x 1 x2
dx
1 1 1 1 2 . 2 1 x 2 dx 2 1 x 2 tan 1 x 2 tan x . 2 1 x 2 1 x2 2
1 x 2 tan 1 x log x 1 x 2
C
Hence,
dx 1 x2 1
f x tan 1 x and A = 1.
7.(C) 8.(BD)
ex
Let
I
Putting
1 ex t 2
x
dx
1 ex
x log t 2 1 , we get :
i.e.
1 t2 I 2 log t 2 1 dt 2 t log t 2 1 2 dt 2 t log t 2 1 2 1 dt 2 2 t 1 t 1
1 ex 1 t 1 x x C 2 t log t 2 1 2t log C 2 x 1 e 4 1 e 2 log 1 ex 1 t 1
1 ex 1
f x 2 x 4 2 x 2 and g x
Hence, 9.(C)
1 ex 1
Putting sin x t in the given integral, we get :
1 t2 1 t2
2
dt
t2 t4
1 t 2 t dt 2
2
t2 t4
2 6 2 1 dt t 6 tan 1 t C 2 2 t t 1 t
sin x 2 sin x
10.(AD)
x 1 x 1
2
Therefore,
2
4
1
6 tan 1 sin x C
dx
1 1 1 dx 3 x2 1 x2 4
1 1 x tan 1 x tan 1 C 3 6 2
A
1 1 and B 3 6
VMC/Solutions/Integral Calculus-1
48
HWT-5/Mathematics
Vidyamandir Classes
Integral Calculus-1 1.(C)
e tan sec x
2
HWT - 6
x sec x sec 2 x tan x sec 2 x tan x sec x dx
e sec x sec x tan x sec x tan x sec x tan x sec 2 x dx
e
sec x
sec x tan x . sec x tan x dx
e
sec x
sec x tan x sec x e
sec x tan x e sec x
2
. sec x tan x sec 2 x dx
sec x
dx
e sec x tan x sec x dx sec x
2
e sec x sec x tan x C 2.(B)
sin8 x cos8 x
1 2 sin
2
3.(A)
sin
2
x cos 2 x
sin
4
x cos 4 x
1 2 sin x cos x 2
ax 2 b
x c 2 x 2 ax 2 b
4.(A)
dx
x cos 2 x
2
dx
2
a b x2
b c ax x
2
cos 2 xdx 2 sin 2 x C 1
dx
2
1 b c 2 ax x
2
b ax x b 1 d ax sin k x c
We have
4e x 6e x 9e x 4e x
4e2 x 6 9e2 x 4
6 A 9e
dxd 9e 4 18Be
Let
4e 2 x 6 A 9e 2 x 4 B .
i.e.
4e2 x
2x
2x
4
2x
Comparing the coefficients of e2x and the constant term, we have 3 35 4 9 A 18B and 6 4 A A and B 2 36 d A 9e2 x 4 B 9e2 x 4 4e x 6e x dx dx dx 9e x 4e x 9e2 x 4
Ax B log 9e2 x 4 C A
Hence,
x x 1 dx 3 t 1 t dt , where t x
6.(A)
1
1
1
3
1 x 2x 1
3 35 ,B and C is any constant. 2 36
5.(B)
2
3 35 x log 9e 2 x 4 C 2 36
dx
1
x 1
1 dx x 1
2
3
1
x3 1 1 1 1 t 1 1 dt log C log 1 x3 3 t 1 t 3 t 3
C
dx
1 dx x 1
x 1 x 1 0 x 1 x 1
log x 1 C
A 1
VMC/Solutions/Integral Calculus-1
49
HWT-5/Mathematics
Vidyamandir Classes 7.(C)
The anti-derivative of f x e x
g x
f x e
x2
2
is given by
dx 2e x
C
2
. . . .(i)
The curve y g x passes through the point (0, 3).
3 3e0 C C 1
Putting C = 1 in (i), we get :
g x 2e x 8.(A)
2
1
We have f x tan 1 xdx C
x tan 1 x
1 x dx C 2 1 x
x tan 1 x
1 2t 2 dt C , where x t 2 2 1 t2
x tan 1 x
t2 1 1
t2 1
dt C
x tan 1 x t tan 1 t C x tan 1 x x tan 1 x C
Since
y f x passes through (0, 2). Therefore, C 2 .
f x x tan 1 x x tan 1 x 2
Hence,
Integral Calculus-1 5.C
tan
1
HWT - 7
1 x dx c 2 1 x
x dx xtan 1 x
= xtan 1 x = xtan 1 x
2t 2
1 2
1 t
t 2 1 1
2
t2 1
c, where x t 2 dt c
= xtan 1 x t tan 1 t c = x 1 tan 1 x
x c
Since y = f (x), passes through (0, 2), c = 2. 7.(B)
8.(C)
f x 1 3x log 3 F 2 7
C 4
F x 0
x 1.
F x Put
cos
dx 2
x 1 tan x
F x
Put
F x x 3x 4
sec 2 x dx 1 tan x
tan x t to get : F x F 0 4
9.(A)
F x x 3x C
4 3cos
dt 2 1 t C 2 1 tan x C 1 t
C=2
dx 2
x 5 sin x 2
tan x t to get F x
sec 2 x dx
4 1 tan x 3 5tan 2
9t
VMC/Solutions/Integral Calculus-1
dt 2
1
2
x
1 1 C tan 1 3t C tan 1 3 tan x C 3 3
50
HWT-5/Mathematics
Vidyamandir Classes 10.(C)
F x
tan x sec 2 x dx tan x
tan x dx sin x .cos x
tan x t to get F x
Put
sec 2 x dx tan x
dt 2 t C 2 tan x C t
F 6 C 4 4
Integral Calculus-1 1.(D)
dx dx 1 1 1 2 2 dx 4 2 2 2 3 x 1 x 4 x 5 x 4 x 1 x 4
I
1 1 x I tan 1 x tan 1 C 3 6 2
2.(C)
f x x 1 x 2 dx
x x 1 dx
4
I
x
2 x4 2 x2 1
2
2 x
5.(D)
4
C
4
2
Put
32
x x 1
x 1 dx 3
1 dt 1 1 1 dt 4 t t 1 4 t t 1
x4 1 1 t log C log 4 C x 1 4 4 t 1
4.(D)
1 2 3 2 1 t C 1 x2 2 3 3
4 x3 dx
1 4
x 4 t to get : I
Put
1 x 2 dx
7 8 C 3 3
f 0
I
1
2 2 x
1 x 2 t to get : f x
Put
3.(B)
HWT - 8
2
1 x
4
1 1 3 5 dx x x 2 1 2 2 4 x x
t to get : I 2
1 2t dt t 4 C t 2
2 x4 2 x2 1 2 x2
C
Integrating by parts we get :
Given integral = 3 x 1 sin x 3 sin x dx 1 2 x cos x 2 cos x dx 3 x 1 sin x 3 cosx 1 2 x cos x 2 sin x 3 x 3 sin x 2 2 x cos x
6.(C)
I
Put
sin 2 x
cos
6
tan x t to get : I t 2 t 2 1 dt
7.(C)
dx tan 2 x sec 2 x . sec 2 x dx tan 2 x 1 tan 2 x . sec 2 x dx
x
t5 t3 C 5 3
tan5 x tan3 x C 5 3
Let e x C1 and e x C2 be the two antiderivatives. Then difference between them is C1 C2 which is fixed and equal to 2.
8.(B)
x sin x
1 cos x dx 2 x sec 1
2
x sin x 2 dx 1 cos x dx
1 x x x 2 xtan 2 tan dx tan dx 2 2 2 2
x x tan C 2
VMC/Solutions/Integral Calculus-1
51
HWT-5/Mathematics
Vidyamandir Classes 9.(C)
tan3 x 3
sec
2
x 1 dx
tan3 x tan x x C 3
10.(C)
I tan 4 x dx tan 2 x . sec 2 x 1 dx
e3 x e x dx I 4 x 2 x e e 1
t
e t to get I
Put
1 1 2 t
1 1 2 t
t 1 dt
x
2
4
t
t2 1
2
1
dt 1 t2
dt . 2
1 t t 1
u
1 Put t u to get I t
du 1
2
tan 1 u tan 1 e x e x
Integral Calculus-1 1.(B)
3 cos x 2 sin x 4 sin x 5 cos x m 4 cos x 5 sin x
23 2 and m 41 41
The given integral is
2.(B)
Differentiate on both sides to get f x
3.(D)
I
Put
1 xx
32
dx
x t to get
cos ec 2 x dx 2 log
5.(D)
I
7.(B)
1
cos
2 xdx x x x3 2
2t dt t t 2
dx
x sin x 4
Put
t 2 u to get I
x
2
x
4
x2 1
12
1 u
dt 4 1 t C 4 1 x C 1 t f x
1 log x and g x tan x 2
4
du
dx
tan 1 u tan 1 tan 2 x
2
x 1 x 1 dx 2
2
1
x3 x 2 1
x
1 t to get I x
Put
t 2 1 u to get I
1 2
x2 1
x 1 x 1 2
2t dt
x
x 12 dx
sin x x
1 t
Put
I
1 tan 4 x
tan x t to get I
I
2 tan x .sec 2 x dx
Put
x4 1
2
3
23 2 4 cos x 5 sin x 23 x 2 dx dx log 4 sin x 5 cos x C 41 41 4 sin x 5 cos x 41 41
.
tan x C
sin 2 x 4
2
I
4.(B)
6.(D)
HWT - 9
2
tan 1 x
2
t dt t 1 2
x 1
x 1
f x tan 2 x
1 1 x x 1 2 dx x 2
1 x x 1
2t dt t 2 1 x4 x2 1 C x
2 x dx 2
1
du u C u
dx
2
1 2
2
2
C
VMC/Solutions/Integral Calculus-1
52
HWT-5/Mathematics
Vidyamandir Classes 8.(A)
x 2 u to get I
Put
x
x dx 2
4x 8
4
du
2
9(C)
2
x2 2 C
1
Put tan 1 x u to get
I etan
1 x 1
dx eu 1 tanu tan 2 u du
x x2 1 x2
eu tanu sec 2 u du eu .tanu C etan
10.(C)
1 2u du 1 u du 2 2 log u 2 4 tan 1 C 2 u2 4 2 2 u 4
1 log x 4 x 8 tan 2
u2
u
I
dx 2 3x x 2
x
dx 2
3x 2
1 x
.x C
dx 2 3 17 x 2 4
3 x 2 C sin 1 2 x 3 C sin 2 17 17 17 3 x 2 4 2 dx
Integral Calculus-1 1.(D)
I
1
HWT - 10
2 sin 2 2 x cos 4 x 1 . sin x . cos x dx dx cos 2 x sin 2 x cot x tan x
1 cos 2 2 x
1 1 t2 1 t2 . dt log t C 2 t 2 2
cos 2 x
x
2.(D)
I
5 4 cos x 51 t 4 1 t , where t tan 2 2 t
3.(B)
I
sin x cos x 2
4.(D)
I
x 1 x
5.(A)
I
dx
2 dt
2
cos x dx
1
dx
2
4
2
cos x sin x cos x sin x . dx sin x cos x
dt 2
cos 2 x 1 log cos 2 x C 2 4
1 2 x tan 1 tan C 3 2 3
9
1 cos x sin x 1 1 dx x log sin x cos x C 2 sin x cos x 2
1 1 1 1 1 x dx tan 1 x tan 1 C 3 x 2 1 x 2 4 3 2 2
fog x cos x dx
t2
2
2
. sin 2 x dx
sin 2 x
1 sin
2
cos x dx . Put sin x t to get x
1 1 1 dt 1 dt t tan t C sin x tan sin x C 1 t2
I
1 t
6.(D)
I
x log
7.(A)
I e sec x . sec3 x sin 2 x cos x sin x sin x . cos x dx e sec x .sec x tan 2 x sec x sec x . tan x tan x dx
8.(C)
9.(C)
2
1
e
ex
e dx
sec x
sec xtan x sec x tan x sec 2 x sec x . tan x dx e sec x sec x tan x C
f x g x f x g x dx f x g x f x g x dx g x f x g x f x dx f x g x g x f x f x e x x 1 x 2 . Since e x 0 for all x R f x 0
10.(B)
1 1 dx log e 1 log e x C x 1 log e x
I
x 1 x 2 0
x 1, 2
1 dx 1 dx 2 2 1 sin x 1 sin x
x
sin 2 x
1 tan 1 2
sec 2 x 1 dx 1 2 tan 2 x
2 tan x C
VMC/Solutions/Integral Calculus-1
53
HWT-5/Mathematics
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