MATHEMATICS

cos x  sin x  1  x

2 x  e sin x  1
Q.1 If dx = n(f(x)) + (A) n +C
e x  sin x  x 2 x  e sin x  1
g(x) + c. Where c is the constant of integration
xe sin x  1  1
(B) n +C
xe sin x  1  1
and f(x) is positive, then g (lim
x )0
(f ( x )) is –
2 xe sin x  1  1
(A) 1 (B) 2 (C) 0 (D) 1/2 [A] (C) n +C
2 xe sin x  1  1

2 xe sin x  1  1
Q.2 Let  g ( x ) dx = F(x), then (D) 2n +C [C]
2 xe sin x  1  1
x
3
g ( x 2 )dx =

e
sec x
1 2 Q.7 Let [secx . tan x . f(x) + (sec x . tan x
[x (F(x))2 –  ( F( x )) dx ]
2
(A)
2 +
1 2
(B) [x F(x2) –  F( x ) 2 d ( x 2 ) ] sec2 x)] dx = esecx . f(x) + C then f(x) may be equal
2 to -
1 2 1
 (F( x )) (A) sec x + tan x (B) sec x – tan x
2
(C) [x F(x) – dx ]
2 2 (C) – x sec x + tan x (D) sec x – x tan x [A]
(D) None of these [B]

dx
x 2  x 2 Q.8  ( x  2) =
 x(x 2  x 2 )3 / 2
7/8
Q.3 dx = .( x  3) 9 / 8
1/ 8 1/ 8
x x 8 x2 5 x2
(A)    c (B)   c
(A) 2
+c (B) 4
+c 5 x 3 8 x 3
1 x 1 x
1/ 8 1/ 8
x 5 x 3 8 x 3
(C) +c (D) None of these [B] (C)    c (D)   c
1 x 8 x2 5 x2
[A]

 (1 – x
2 x dx f (x)
Q.4 = , then f(x) is - x 2
2
) x 4
–1 x2 –1 6 0
 
Q.9 Let A = 1 –5 1 and B =
(A) x 4
1 (B) x 4
–1  2 0 x
 
(C) – 1– x4 (D) None of these [B]

4 0 0
2x  1 0 0
Q.5 If x 4
 2x 3  x 2  1
dx = A n 

1 .
8
0 0 
If a function is defined as f(x) = tr (AB), then
x2  x 1
+ C then the value of A can be -
1 x2  x
3dx
(A) 1 (B) 1/2 (C) – 1/2 (D) – 1 [C]
 f (x) is equal to

( x cos x  1)dx
Q.6  2 x 3 e sin x  x 2
is equal to –

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 1 2x  1 cos ec 2 x  2005 A( x )
(A)
4
n
2x  5
C (B) Q.11 If
 cos 2005
x
dx = –
( B( x )) 2005
+ c, then number of solutions of the equation
1 2x  5
n C
4 2x  1 A( x )
={x} in [0, 2] is (where {.} represents
B( x )
1 1  2x
(C) n C (D) fractional part function)
3 2x  5
(A) 0 (B) 1 (C) 2 (D) 3
1 1  2x
n C cos ec 2 x  2005
3 2x  3
[A]
Sol.[A]
 cos 2005 x
dx

Sol.  cos ec x  (cos x ) 2005

2
= dx – 2005
4 x 2 6 0
  1
AB =  4
 8
–5
0
8
8x 
 (cos x ) 2005 dx = I1 – I2
 
Applying by parts on I1, we get
 f(x) = tr(AB) = 4x2 + 8x – 5
3 dx cos ec 2 x  2005
  4x 2
 8x  5
=
 cos 2005 x
dx

3 dx cot x
=  C
 (2x  5)(2x  1) (cos x ) 2005
A(x) = cotx and B(x) = cos x
1  1 1 
=
2     dx
 (2 x  1) ( 2 x  5)  
A( x )
= cosec x = {x} for x [0, 2] the
B( x )
1 2x  1
= n C equation has no solution as clear from the
4 2x  5
graph
sec x . cos ecx (,1)
Q.10  2 cot x  sec x cos ecx dx is equal to-

1 (1,0)
(A) n | sec2x + tan 2x | + C
2
(B) n | secx + cosec x | + C
(C) n | sec x + tan x | + C Q.12 Let x2 n –1, n  N, then
1
(D) n | sec x + cosec x | + C [A] 2 sin( x 2  1) – sin 2( x 2  1)
2  x
2 sin( x 2  1)  sin 2( x 2  1)
dx is
sec x cos ecx
Sol. I = 
2 cot x  sec x cos ecx
dx =
equal to:
1
dx (A) n sec( x 2  1) + C
 2 cos 2
x 1
2

 x2 1
1 (B) n sec  +C
=  sec 2 x dx =
2
n | sec 2x + tan 2x | +  2 

C 1
(C) n |sec(x2 + 1)| + C
2

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 1 2 2x
(D) n +C f(x) = s =
1  2x 2
2
2 sec( x  1)

1 2x
Sol.[B] I =
2 so  1  2x 2 dx put 1 – 2x2 =

t
2 2
2 sin( x  1) – sin 2( x  1)
 2x 2 sin( x 2  1)  sin 2( x 2  1)
dx – 4x dx = dt
2x dx = – 1/2 dt
x2 + 1 = t  2x dx = dt 1 dt 1
=–
2  t
= –
2
log (t) + c
1 2 sin t – sin 2 t
I=
2  2 sin t  sin 2 t
dt
=–
1
log (1 – 2x2) + c
2
1 2 – 2 cos t 1 t
=
2  2  2 cos t
dt =
2  tan 2 dt = log 
 1
 +c
2 
 1  2x 
t
n sec
1 2  x2 1
= + c = n sec  +  3x  4 
2 1  2 
Q.14 If f   = x + 2 then
 3x  4 
 f ( x )dx is
2
equal to
c
3x  4
(A) ex+2 ln +c
3x  4
Q.13 If f(x) = nlim

(2x + 4x3 +……..+ 2nx 2n–1)
8 2
(0 < x < 1) then  f ( x ) dx is equal to (B) 
3
ln |(1 – x)| +
3
x+c
 
1
(A) log   +c
8 x
2  (C) ln |x – 1| + +c
 1 x 
3 3
(B) log 1  2x 2 + c
(D) None of these
 1 
(D) log   +c
 3x  4
2 Sol.[B] =t
 1  2x 
3x  4
(D) None of these
3x – 4 = 3xt + 4t

Sol.[C] Solve series 4t  4

x=
3(1  t )
s = 2x + 4x3 +…+ 2nx2n–1  Total term (n)
4t  4
 G.P. f(t) = +2
3(1  t )
[(2 x 2 ) n  1]
s = 2(x)  first term = 2x 4x  4 4( x  1)  8
(2 x 2  1) f(x) = +2= +2
3(1  x ) 3(1  x )
 C.R. = 2x2
4 8 2 8
f(x) = 2 – – = 
( 2 x ) [ 2 n x 2n  1] 0  x 1 3 3( x  1) 3 3( x  1)
s = (2x) , n
( 2 x 2  1)
2 8
x 02n  f ( x )dx =
3
x–
3
ln |x – 1| + c

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 (sin x ) n  (sin x )  n | e ibx | = cos 2 bx  sin 2 bx =1
Q.15 If f(x) = nlim
 (sin x ) n  (sin x )  n
,0<x<

/2, Q.17
 (cos x )[f ( x )]
3
n  N, then dx is equal to
cos 2 x sin 3 xdx
(A) +c (B) sin 2x + c
2  (cos 4
x  3 cos 2 x  1) tan 1 (sec x  cos x )
(C) – sinx + c (D) None
Sol.[C] 0 < x < /2 , 0 < sinx < 1 =
lim (sin x)n  0 (A) tan–1 (sec x + cos x) + c
n 
(B) log tan–1 (sec x + cos x) + c
(sin x ) 2n  1
f(x) = nlim
 =–1 (C) 1/(sec x + cos x) + c
(sin x ) 2 n  1 (D) None of these

 cos x ( 1)dx = – sin x + c Sol.[B] Put tan–1 (sec x + cos x) = t

1
(sec x tan x – sin x) dx =
ax ax 1  (sec x  cos x ) 2
Q.16 If P =  e cos bx dx and Q =  e sin
bx dx, then (P2 + Q2) is equal to (neglect dt
integration constant)
cos 2 x  1
eax e bx 2 2 2 sin x 1
2
(A) B) cos x  (1  cos x )  cos x 
a 2  b2 a 2  b2
dx = dt
e 2ax e (a  b) x
(C) (D)
a 2  b2 a 2  b2 (1  cos 2 x )
cos 2 x sin x
Sol.[C] P + iQ = e
ax
cos bxdx + i cos 2 x dx = dt
2 4 2
cos x  1  cos x  2 cos x
ax
e sin bx dx
sin 3 x
dx = dt
P + iQ =  e ax (cos bx  i sin bx ) dx cos 4 x  3 cos 2 x  1
dt
P + iQ =  (e ax eibx ) dx I=  t
 log | t | + c

P + iQ =  e (ax ibx ) dx  log |tan–1 (sec x + cos x)| + c

e ( a  ib ) x
P + iQ =
(a  ib) Q.18 The indefinite integral of (12 sin x + 5 cos x) –1
is, for any arbitrary constant -
e ax eibx
(P + iQ) =
(a  ib) x 1 5 
(A) 1/13 log tan  tan 1  +c
Take modulus on both side 2 2 12 
x 1 12 
e ax (B) 1/13 log tan  tan 1  +c
P 2
Q 2 = 2 2 5 
a 2  b2
1 5 
2ax
(C) 1/13 log tan  x  tan  +c
e  12 
P2 + Q2 =
(a  b 2 )
2
1 12 
(D) 1/13 log tan  x  tan  +c
 5 
 eibx = cos bx + i sin bx

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 5 5
Sol.[A] With = tan–1 , sin  = and cos  =
12 13 (/2, 1)

12
13
dx dx (1, 0)
 12 sin x  5 cos x
=  13 sin( x  )
1 x
sec 2
1
=
13  2
x
2 dx
dx
tan
2
Q.20 If I =  cos5 x  sin 5 x , then I equals
1 x 1 1 5  2t 1
= log tan  tan  +c (A) 2 log + log
13  2 2 12  2 t (2  b 2 ) b 3

bt
cos ec 2 x  2005
Q.19 If
 cos 2005 x
dx = – bt

1 2
t
+ tan–1  
2 3
(2  b ) b b
A( x )
+ C,
( B( x )) 2005 +C

then number of solutions of the equation 5t 1 bt

(B) 5 log + log
5 t 2  b2 bt
A( x )
= {x} in [0, 2] is (where {.} 2 t
B( x ) + 3
tan–1   + C
b b
represents fractional part function) -
5 t t
(A) 0 (B) 1 (C) 2 (D) 3 (C) 5 log + sin–1   + C
5t b
cosec 2 x  2005
Sol.[A]
 cos 2005 x
dx
(D) 5 log
2 t
2 1
t  2 
+ sin–1   +  3  tan–1
b b 
 cosec x.(cos x ) 2005
2
= dx
1 t
– 2005  (cos x ) 2005 dx = I1 – I2   +C
b
Applying by parts on I1, we get
where t = sin x – cos x and b = 51/4
2
cosec x  2005 cot x
 cos 2005
x
dx = –
cos 2005
x
+c Sol.[A] cos5 x + sin5 x = (cos x + sin x) [cos 4 x –cos3x
sin x + cos2 x sin2 x –cos x sin3 x + sin4 x]
= (cos x + sin x) [(cos 2x + sin2x)2 – cos2 x sin2 x
 A(x) = cot x and B(x) = cos x – cos x sin x (cos2 x + sin2 x)]
A( x ) = (cos x + sin x) (1 –cos x sin x – cos2 x sin2 x)
 = cosec x = {x} for x  [0, 2] the
B( x ) I=
equation has no solution as clear from the graph

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 dx
Q.22 If I = x 1 x3
, then I equals:
sin x  cos x
 (1  2 sin x cos x ) (1  cos x sin x – cos 2 x sin 2 x ) (A)
1
log
1  x3 1
+C
3 1 x3 1
dx
1 1 x3 1
Put sin x – cos x = t, so that (cos x + sin x) dx = (B) log +C
3 1 x3 1
dt, and t2 = 1 –2 sin x cos x 2
(C) log |1 – x3| + C
Thus 3
dt 1
  2 
(D)
3
log x 3/ 2  1 x3 +C
I= (2  t ) 1  t  1  1 ( t 2  1) 2 
2

 2 4  x 2 dx
dt
Sol.[B] Write I =  x3 1 x3
and 1 –x3 = t2, so that
=4  (2  t 2 ) (5  t 4 )
–3x2 dx = 2t dt and
=4 ( 2 / 3) t dt 2 dt
 1 1 1 1 1 1 1 
I=  (1  t ) t2 =–
3  1 t 2
  2  t 2  2  5 2 5 5 t 2

2 5 2 5

5t  2
2 1 1 t
= –   log + C = 3 log
dt 3 2 1 t

2t 1 1
= 2 log log
2 t 2 5 5 3/ 4
1 x3 1
+C
1 x3 1
51 / 4  t 1 2  t 
+ tan–1  + x5
5 1/ 4
t 2 5 5 3/ 4
 51 / 4  Q.23 If I =  1 x3
dx, then I is equal to:

C
2 2
where t = sin x – cos x (A) (1 + x3)5/2 + (1+ x3)3/2 + C
9 3
Q.21 If I =  tan x  
cot x dx, then I equals: 2 2
(B) (1 + x3)3/2 – (1 + x3)1/2 + C
(A) –1 9 3
2 sin (sin x + cos x) + C

(B) –1 (C) log x  1 x3 +C

2 cos (sin x –cos x) + C

(C) –1 (D) x2 log (1 + x3) + C

2 sin (sin x –cos x) + C
–1
Sol.[B] Put 1 + x3 = t2, 3x2 dx = 2t dt, so that
(D) 2 cos (sin x + cos x) + C
( t 2  1) 2 2  t 3 
 t + C
Sol.[C] I = 
sin x  cos x
dx
I=  t
.
3
t dt =
3  3 

cos x sin x
Put sin x – cos x = t, so that (sin x + cos x) dx = dt 2 2
= (1+ x3)3/2 – (1 + x3)1/2 + C
9 3
and 1 – 2 sin x cos x = t2
1 1 x
I= 2 
dt
= –1
2 sin (t) + C
Q.24 If I = x 1 x
dx, then I equals:
1 t2

= –1 (A) log |x| + log 1 1 x2 + sin–1 x +C

2 sin (sin x – cos x) + C
(B) log |x| – log 1 1 x2 + tan–1 x + C
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 (C) log |x| – log 1 1 x2 – sin–1 x + C  2t 
 I = sin–1   + C = sin–1

(D) log 1 1 x 2
– log |x| + cos–1 x + C
 2x     
1 x   + C
1   
Sol.[C] I = x 1 x2
dx
x 2  1 dx, then I equals
dx dx Q.26 If I =

= x 1 x2
–  1 x2
x4
1 ( x 2  1) 3 / 2
(A) – +C
1 3 x3
In the first integral, put x = , so that
t (B) x3 (x2 + 1)–1/2 + C
( 1 / t 2 ) dt x 2 1 + C
I=  1 1
–1
– sin x = – 
dt (C)
1 2 t 2 1 x2
t t
1 ( x 2  1) 3 / 2
–1
– sin x (D) – +C
3 x2
–1
= – log | t + t 2  1 | – sin x + C
Sol.[A] Put x = tan  so that x 2  1 = sec , dx = sec
2

2
1 1 x
= – log  – sin–1 x + C = log |x| – d
x x
sec  sec 2  cos 
log |1 + 1 x 2 | – sin x + C–1 I =  tan  4
d =  sin 4  d
1 1
dx =– +C
Q.25 If I =  ( x   ) (  x )
. (< ) then 3 sin 3 

1 sec 3  1 ( x 2  1) 3 / 2
value of I is: =– +C=– +C
3 tan 3  3 x3
 2x     
(A) sin–1   + C
  
dx
(B) sin –1
 x   
 
Q.27 If I =  sec x  cosec x ' , then I equals :
  
1  1 
 2x      (A)  cos x  sin x  log(cos ec x – cos x ) 
(C) sin  –  2  2 
  
+C
(D) None of these
1  1 
1 (B)  sin x  cos x  log | cosec x  cot x | 
Sol.[A] Put t = (x –  + x – ) 2  2 
2
1 +C
=x– ( + ), so that
2 1  1 
(C)  sin x  cos x  log | cosec x  cos x |  +
 1  2  2 
(x – ) ( – x) =  t  (  )   
 2 
C
 1 
  2 (  )  t  (D) None of these
 
sin x cos x
=
1
( – )2 – t2
Sol.[D] I =  sin x  cos x dx
4

Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 INDEFINITE INTEGRATION 7
 1 (sin x  cos x ) 2  1 = –
=
2  sin x  cos x
dx

  dn (  x )  
=
1 
 sin x  cos x 
1 
 

n (  x ) x dx  
 dx   xdx  dx 
2 2 sin( x   / 4) 
dx  x2 1 x2 
1 1 =–  n (  x )   . dx 
= [sin x – cos x] – log |cosec (x + /4)  2 x 2 
2 2
– cot (x + /4)| + C x2 x2 
=–  n ( x )  +c
 2 4 
Q.28  (x – 10C1 x2 + 10C2x3 –10C3x4 .....+ 10C10x11)
dx equals: 1 1
= x |x| n |x| – x |x| + c ( x < 0)
(1  x ) 11
(1  x ) 10 2 4
(A) – +c
11 10
12
when, x > 0; I =  xnx dx
(1  x ) (1  x )11
(B) – +c  dnx 
12 11 = nx  x dx –   dx  xdx dx
10
(1  x ) (1  x )11
(C) – +c
10 11 x2 x2
11
= nx – +c
(1  x ) (1  x )12 2 4
(D) – +c
11 12 1 1
= x |x| n |x| – x |x| + c (x > 0)
2 4

Sol.[B] Given expression Q.30 I (cos x)4 dx = Ax + B sin 2x + C sin4x then

10 10 2 10 10
= x (1– C1 x + C2x + .....+ C10x )
{A, B, C} equals-
= x(1– x)10
3 1 1  3 1 1 
(A)  , ,  (B)  , , 
 x (1  x )
10
I= dx  8 4 32  8 2 6 
Let I –x = t – dx = dt
3 1 1  3 1 1 
(C)  , ,  (D)  , , 
I=–  (1  t ) t 10 dt  4 4 16   8 4 32 
2
 t 11 t 12  1  cos 2 x 
=
11
+
12
+c Sol.[D] I =   2  dx
1
=  (1  cos 2 2 x  2 cos 2 x ) dx
Q.29  |x| ln |x| dx equals (x 0) 4
1  1  cos 4x 
(A)
1
x |x| ln |x| –
1
x |x| + c
=
4 
 1 
2
 2 cos 2x  dx

2 4
1
1 1 =
8
 (3  cos 4 x  4 cos 2x ) dx
(B) x |x| ln |x| + x |x| + c
2 4 3x sin 4 x 4 sin 2 x
= + +
1 2 1 8 32 8 2
(C) x ln |x| – x |x| + c
2 4 dx 3
(D) None of these
Q.31  x 3 ( x n  1) equals:
Sol.[A] If x < 0, I = –  xn (  x ) dx

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  xn   n  3
(A)
3
ln  n  (B) 1 ln  x  (A) x2/3 + 6 tan–1 6
x +c
  x n 1  2
n  x 1  n  
3
(B) x2/3 + 6 tan–1 x +c
3  x n 1  x n 1  2
(C) ln  
n  (D) 3n ln

 xn 

n  x    3
(C) x2/3 + tan–1 x + c
2
dx 3 x 2 dx 3
Sol.[A]  x 3 ( x n  1) = 3  x 3 ( x n  1) (D)
2
x2/3 + 6 tan–1 x1/3 + c

dx Sol.[A] Let t6 = x
=3  x ( x n  1) then 6t5 dt = dx

x n 1 dt ( t 6  t 4  t )6 t 5 dt t5  t3 1
=3
x n ( x n  1)
dx =
3
 t ( t  1)
I=  6
t (1  t ) 2 =6  (1  t 2 )
dt
n

3 ( t  1)  t 3 1 1  t 3 ( t 2  1)  1
=
n
 t ( t  1)
dt =
n
  t  t  1  dt =6  (1  t 2 )
dt

3 1
t  1 t2
3
= (nt – n (t + 1)) + c =6 dt  6 dt
n
3  t  6t 4 3 4
= n  +C = + 6 tan–1 t = t + 6 tan–1 t + c
n  t 1  4 2

Q.32  ( cot x + tan x ) dx. equals:

=
3 2/3
x + 6 tan–1 6
x +c
–1
2
(A) 2 . sin (sin x – cos x)
dx
(B) –1
2 sin (sin x + cos x) Q.34  ( x 2  1) ( x 2  4) equals:
(C) 2 sin–1 (sin x – cos x)
1 1
(D) None of these (A) tan–1 x + tan–1 x + c
3 6
 cos x sin x 
Sol.[A] I =   sin x

cos x
 dx


(B)
1
tan–1 x –
1
tan–1
x
+c
3 6 2
cos x  sin x x
=  sin x cos x
dx (C) tan–1 x –tan–1 +c
2
cos x  sin x 3 x
= 2  sin 2 x
dx (D) 3 tan–1 x –
2
tan–1
2
+c

put t = sin x – cos x dx

dt = (cos x + sin x) dx
Sol.[B]  ( x 2  1) ( x 2  4)
also t2 = 1 –sin 2x ( x 2  4)  ( x 2  1)
1
dt
=
3  ( x 2  1) ( x 2  4)
dx
I= 2  1 t2
= –1
2 sin t
1 dx 1 dx
= –1
2 sin (sin x –cos x) + c
=
3  x2 1 – 3  x2  4
3
x  x2  6 x 1 1 x
Q.33  x (1  3 x )
dx equals: =
3
tan–1 x –
6
tan–1
2
+c

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 Questions Add (24–6-09) e x dx
Q.37  cos hx  sin hx
is
sec x. cos ecx
Q.35  2 cot x  sec x cos ecx dx is equal to
(A)
1 2x
e +c (B) x + c
2
1
(A) n | sec2x + tan2x | + C (C) log (cot hx) + c (D) log (tan hx) + c
2
(B) n | sec x + cosec x | + C Sol.[B] sin hx = (e – e )/ 2, cos hx = (ex + e–x)/2
x –x

(C)n | sec x + tan x | + C

1
(D) n | sec x + cosec x | + C ( x  1)dx
2 Q.38  (x  1) x ( x 2  x  1)
is
Sol.[A]
sec x cos ecx dx 1
 2 cot x  sec x cos ecx dx   2 cos 2
x 1
(A) tan 1 x
x
1 + c

1
=  sec 2x dx  2 n | sec2x + tan2x | + C (B) 2 tan 1 x 
1
x
1 + 

Q.36 Let x2 n – 1, n  N then

1
(C) 2 tan –1 x 1 + c
2 sin ( x 2  1) – sin 2 ( x 2  1)
x dx x
2 sin ( x 2  1)  sin 2 ( x 2  1)
1 1
equals (D) tan 1 x 1 + c
2 x
1
(A) n sec ( x 2  1) +C ( x  1) dx
2
 1
 x 2 1  Sol.[C] I = ( x  1) x x 
1
(B) n sec  
 +C x
 2  ( x 2  1) dx
(C)
1
n | sec(x2 + 1) | + C
2 1
= ( x  2 x  1) x x   1
2 x
1  1 
(D) n
2
+C 1  2  dx
 x 

2 sec ( x 2  1)

1  1 
=  x   2 x   1
1
Sol.[B]
2  x  x
put x + 1/x + 1 = t2
2 sin ( x 2  1) – sin 2 ( x 2  1)
 2x 2 sin ( x 2  1)  sin 2 ( x 2  1)
 sec 2 x  1 
dx
Put x2 + 1 = t 2xdx = dt
Q.39  x sec 2x  1  dx =
1 2 sin t – sin 2 t x2
=
2  2 sin t  sin 2 t
dt (A) x tan x – log |sec x| 
2
c

1 2 – 2 cos t
=
2  2  2 cos t
dt
(B) x tan x – log |sec x| +
x2
c
2
1
1 t n | sec t / 2 |  C

2  tan
2
dt = 2
(C) x tan x + log |sec x| 
x2
c
1/ 2 2
x 2
1 
= n sec   +C x2
 2 

(D) x tan x + log |sec x| + c
2
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  1  cos 2 x  (C) 2.4 (D) 2.5
I =   1  cos 2 x  dx =  tan x dx
x x 2
dx
Sol.[A]
use integration by parts Sol.[C] f(x) =  sin1 / 2 x
. cos 4 x
1/ 2
cos x

x r 2r
Q.40  dx = sec 4 x
r0
r f(x) =  tan x
dx =

e 2x
(A) ex + c (B) +c
2 (1  tan 2 x ) sec 2 x dx
e 2x
 tan x
(C) +c (D) 2e2x + c 
4
1 t2
  (t
1 / 2
 f(x) = dt =  t 3 / 2 ) dt
x r 2r
Sol.[B] 
r 0
r
= e2x
t

f(x) = 2t1/2 + 2/5 t5/2 = 2 tan x + 2/5 (tan x)5/2

1– x 2  1 x 2
Q.41  dx = 
1– x4 f  – f(0) = 2 + 2/5 = 2.4
4
(A) log ( x  1  x 2 ) + sin–1 x + c Q.44  1  cos ecx dx =
–1 –1
(B) sin h x – sin x + c –1
(A) ± sin (tan x – sec x) + c
(C) cos–1 x – sin–1 x + c (B) 2sin–1 (cos x) + c
(D) tan–1 x + sin–1 x + c   x x
(C) sin–1  cos  sin  + c
1– x2 1 x 2  2 2
Sol.[A]  1– x4

1– x 4
dx
 x x
(D) ± 2 sin–1  sin  cos  + c
dx dx  2 2
=  1 x2
  1– x2
Sol.[D] I =  1  cos ecx dx

e 3x  e x 1  sin x
Q.42 e 4x
–e 2x
1
dx = I =  sin x
dx = ±

(A) tan–1 (ex – e–x) + c

(B) tan–1 (ex + e–x) + c
(C) 2 tan–1 (ex – e–x) + c x x
sin  cos
(D) 2 tan–1 (ex + e–x) + c   2
x
2
x
dx
Sol.[A] Put ex = t 2 sin cos
2 2
 1 
2 1  2  dt x x
(t  1) dt  t 
I = t 4
–t 2
1
=
 t 2 – 1 2
1
I=± 
sin
2
 cos
2
2 dx
t  x x
1   sin  cos 
1  2 2
put t – =z
t
x x  x x
Put sin – cos = t,  cos  sin  dx
2 2  2 2
dx 
Q.43 If f(x) =  sin1/ 2 x cos7 / 2 x then f   –
4 = 2dt
f(0) is equal to
(A) 2.2 (B) 2.3
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 2dt 1
I= ±  1 t 2
= ± 2 sin–1(t) (C)
2
(sin x – cos x + x) + c

(D) None of these

 x x
= ± 2 sin–1  sin  cos  + c
2 2 sinx cosxdx
 Sol.[B] I =  1  sinx  cosx 

x 2x  sin x
e  sec x  tan x  1 dx
 1  .dx is equal to
Q.45  tan x 
 (1  x 2 ) 2 
 1 1  sin x (1  tan x  sec x )
(A) ex  cot

x  +c
1 x2 
I=  (1  tan x ) 2  sec 2 x
dx

 1 1 
sin x (1  tan x  sec x )dx
(B) ex  tan x   +c
 1 x2    2 tan x
 1 1 
(C) ex  tan x   +c 1
 1 x2  I=
2
 cos x (1  tan x  sec x ) dx
(D) ex tan–1 x + c
 2x  1
Sol.[C]  e  tan x 
x 1  dx


2
 (cosx + sin x – 1) dx
 (1  x 2 ) 2 
1
= [sin x – cos x – x] + c
 1  2
 tan 1 x  
 1 x2 
= e Q.47 If f(x) = tan–1 x + n 1  x – n 1  x . The
x
dx –
  integral of 1/2 f ' (x) with respect to x4 is -
f (x) f ' (x) 4
(A) e  x  c (B) – n (1 – x4) + c
(C) e 1 x 4  c (D) n(1 + x2) + c
 1 2x 
 
 1  x 2  (1  x 2 ) 2  Sol.[B] f(x) = tan–1 x + n 1  x – n 1  x
 
e
x

  1 1 1
f '(x) = + +
1 x 2 2(1  x ) 2(1  x )
g(x ) g ' (x)
1 1 2
1 =  
= ex tan–1(x) – ex +c 1 x 2
1 x 2
(1  x 4 )
2
1 x
1 1
f ' (x) =
2 1 x4
 1 
1 1 dx 4
= ex  tan x  +c
 1 x  2
2
f ' ( x )dx 4 
1 x4 
dt
– 
t
– nt + c 1 – x4 = t
dx
Q.46  tan x  cot x  sec x  cos ecx
= – n (1 – x4) + c – dx4 = dt

1 cos 3 x
(A)
2
(sin x – cos x) + c Q.48
 sin 11 x
dx is equal to -

1 2 2
(B) (sin x – cos x – x) + c (A) cot3/2 x – cot9/2 x + c
2 5 9

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 2 2 ax
(B)
5
cot5/2 x –
9
cot9/2 x + c Q.55  ax
dx is equal to-

2 2 (A) sin–1 (x/a) – a 2  x 2  c

(C) cot5/2 x – cot3/2 x + c
5 9 (B) cos–1 (x/a) – a 2  x 2  c
(D) None of these (C) a sin–1 (x/a) – a 2  x 2  c
Sol.[B] Since sum of powers of (D) a cos–1 (x/a) – a 2  x 2  c [C]
3  11 
cos x and sin x = +   = – 4, x
2  2  Q.56  xe dx is equal to-
we will put tan x = t (A) (x + 1) e–x + c (B) (x – 1) e–x + c
(C) – (x + 1) e–x + c (D) (1 – x) e–x + c [C]
4  5 sin x
Q.49  cos 2 x
dx equals-
Q.57  (log x )
2
dx equals-
(A) 4 tan x – sec x + c (B) 4 tan x + 5 sec x + c (A) (x log x)2 – 2x log x + 2x + c
(C) 9 tan x + c (D) None of these [B]
(B) x (log x)2 – 2x log x + 2x + c
(C) x (log x)2 + 2x log x + 2x + c
1  cos 2 x
Q.50  1  cos 2x dx equals- (D) None of these [B]
x  sin x  cos x 
(A) tan x + x + c (B) tan x – x + c Q.58 e 
 cos 2 x
 dx is equal to-

(C) sin x – x + c (D) sin x + x + c [B]
(A) ex cos x + c (B) ex sec x + c
x x 2
(a  b ) (C) ex sin x + c
Q.51  a bx x
dx equals- (D) None of these [B]

x
2
(A) (a/b)x + 2x + c (B) (b/a)x + 2x + c Q.59 tan 3 x 3 sec 2 x 3 dx is equal to-

(C) (a/b)x – 2x + c (D) None of these [D] 1 1

(A) tan4 x3 + c (B) tan4 x3 + c
4 8
3x 2 1
Q.52  x 6  1 dx equals- (C)
12
tan4 x3 + c (D) None of these [C]

(A) log (x6 + 1) + c (B) tan–1 (x3) + c 3x  1

(C) 3 tan–1 (x3) + c (D) 3 tan–1 (x3/3) + c [B] Q.60  2x 2  2x  3 dx equals-
1 5  2x  1 
tan (log x ) (A) log(2x2–2x+3)– tan–1  +
Q.53 The value of  is- 4 2  5 
x C
(A) log cos (log x) + C
3 5  2x  1 
(B) log sin (log x ) + C (B) log(2x2–2x +3)+ tan–1  
(C) log sec (log x ) + C 4 2  5 
(D) log cosec (log x) + C [C] +C
3 5  4x  2 
dx (C) log(2x2–2x+3) + tan–1  
Q.54  ex  ex equals-
+C
4 2  5 

(A) log (ex + e–x) + c (D) None of these [B]

(B) log (ex – e–x) +c
x3 – x – 2
(C) tan–1 (ex) +c Q.61  (1 – x 2 )
dx =
(D) tan–1 (e–x) +c [C]
 x 1 x2
(A) log   – +c
 x 1  2
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  x 1  x 2 dx
(B) log 
 x 1
+
2
+c Q.66  3  sin 2 x equals-

 x 1 x2 1  2 tan x 
(C) log  + +c (A) tan–1  +c
 x 1  2 2 3  3 
 x 1  x 2 1  tan x 
(D) log  – +c [D] (B) tan–1   + c
 x 1 2 3  3 
x
Q.62  ( x 2  1)( x 2  2) dx equals- (C)
1
tan–1 (2 tan x) + c
2
1  x2 1  (D) None of these [A]
(A) log  2 +c

2 x 2 sin x
x 2 2
Q.67 The value of  sin x  cos x dx equals-
1
(B) log  2 +c

2  x  1  1 1
(A) x+ log ( sin x – cos x ) + C
2 2
 x2 1 
(C) log  2 +c
 1 1
x 2 (B) x– log ( sin x – cos x ) + C
2 2
 x2  2  (C) x + log (sin x + cos x) + C
(D) log  2 +c
 [A]
 x 1  (D) None of these [A]

x 2 1
Q.63 The value of  x 4 1 dx equals- dx
Q.68  a sin x  b cos x equals-
1  x  2x  1 
2
(A) log  2  +C

1  1 1 b  
2 2  x  2x  1  (A) log  tan  x  tan  + C
a b2

2 2 a 
1  x 2  2x  1  1
(B) log  2  +C
 (B) log
2 2  x  2x  1  a  b2
2

1 x 2 1   1 b  
(C) tan–1 +C  tan  x  tan  + C
2 2x   a 
(D) None of these [A] 
1 1 1 b  
dx (C) log  tan  x  tan  +
Q.64  equals- a 2  b2  2 a 
5x  6  x 2 C
(A) sin–1 (2x + 5) + c (D) None of these [A]
(B) cos–1 (2x + 5) + c 1
(C) sin–1 (2x – 5) + c Q.69 Evaluate  ( x  3) x 1
dx-
(D) None of these [C]
1 x 1  2
(A) – log +C
2x  3 2 x 1  2
Q.65  dx is equal to -
x2 1 1 x 1  2
(B) log +C
2 x 1  2
(A) 2 x 2  1 + 3 n | x  x2 1 | + c
1 x 1  2
(B) (C) log –C
x 2  1 + 3 n | x  x2 1 | + c 3 x 1  2

(C) 2 (D) None of these [B]

x 2  1 + 3 n | x  x2 1 | + c

(D) None of these [A]

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 1 1 1 x2  2x
Q.70 Evaluate  ( x 2  4) x 1
dx (D) – log – [D]
2 1 x 2
 2x
1 x 1  3 1
(A) n – tan–1 ( log |x + 1 x2 | + C
4 3 x 1  3 4
x  1 ) +c
1 1 dx
(B) – tan–1( x  1 )+ n Q.73  2
=
4 4 x x  2x  2
x 1  3
+c (A) loge |x + 1 + x 2  2x  2 | +
x 1  3
1 1 2
(C) – tan–1( x  1 )+ n +C
4 3 4 x  2  x 2  2x  2
x 1 1
+c (B) loge |x + 1 + x 2  2x  2 | –
x 1 1
(D) None of these [C] 2
+C
x  2  x 2  2x  2
1
Q.71 Evaluate  dx (C) loge |x + 1 – x 2  2x  2 | +
( x  1) x2  4
2
1 1 1 x2  4 +C
(A) – log   +
5 x 1 5 5 ( x  1) 2 x  2  x 2  2x  2

C (D) None of these [A]

1 1 1 x2  4 dx
(B)
5
log  
x 1 5 5 ( x  1) 2 +C Q.74  1 x 2 1
=

2 1 x
(C)
1
log
1 1
 
x2  4
+C (A) 1  1  x – 2 tan–1 +C
5 x 1 5 5 ( x  1) 2 x 1 x
2 1 x
1 1 1 x2  4 (B) 1  1  x – 2 tan–1 +C
(D) log   – C x 1 x
5 x 1 5 5 ( x  1) 2
2 1 x
[A] (C) 1  1  x + 2 tan–1 +C
x 1 x

1  x 2 dx 2 1 x
(D) 1  1  x + 2 tan–1
Q.72 Evaluate  1 x2 x 1 x
+C [C]

15
1 1 x2  2x x 
 1 x 2 

(A) – log + Q.75   dx =
2 1 x2  2x 
1 x2
log |x + 1 x2 | + C 15 15
x  1  x 2   x  1 x2 
1 2 (A) 
  + C (B)  

1 x  2x
(B) log + 30 15
2 1 x 2
 2x
+C
log |x + 1 x 2 |–C 15
x  1  x 2 

1 (C)   +C (D)
1 x2  2x
(C) – log – 30 x
2 1 x2  2x
15
log |x –  x  1 x2 
1 x2 | – C  
+ C [B]
 
15 x

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Q.76 If x = f  (t) cos t + f  (t) sin t, y = – f  (t) sin t + 1
1/ 2 (A) [f(x)]2 (B) [f(x)]3
 dx  2  dy  2  2
f t) cos t, then    
 dt 
 
 dt  
dt is
1
(C) [f(x)]3 (D) [f(x)]2 [A]
equal to - 3
(A) f (t) + f  (t) + c (B) f (t) + f  (t) + c Q.81 x
n
log x dx equals -
(C) f(t) + f  (t) + c (D) f  (t) – f  (t) + c [C]
x n 1 log x  1 
(A)   +c
n 1  n  1
 x 
e sin    dx is equal to –
x/2
Q.77
2 4 x n 1 log x  2 
(B)   +c
n 1  n  1
(A) ex/2 sin x/2 + c
(B) ex/2 cos x/2 + c x n 1 2 log x – 1 
(C)   +c
(C) 2 ex/2 sin x/2 + c n 1  n 1 
(D) 2 ex/2 cos x/2 + c [C] x n 1 log x – 1 
(D)   +c [D]
n 1  n 1 
dx
Q.78  sin( x – a ) cos(x – b) is equal to - log( x  1  x 2 )
sin( x – a )
Q.82  1 x2
dx equals-
(A) cos (a – b) log +c
cos( x – b)
1
sin( x – a ) (A) [log(x+ 1  x 2 )]2 + c
(B) sec (a – b) log +c 2
cos( x – b)
(B) log (x+ 1  x 2 ) + c
cos(x – a )
(C) sin (a – b) log +c (C) log [(x+ 1  x 2 )]2 + c
sin( x – b)
cos( x – a ) (D) None of these [A]
(D) cosec (a – b) log +c [B]
sin( x – b) d2v d 2u
Q.83
 u
dx 2
dx –  v
dx 2
dx is equal to -

1
Q.79 If  ( x –  )( – x )
dx ( < x < ) is equal
(A) uv + c (B) u
dv
–v
du
+c
dx dx
to -
du dv dv du
(i) 2sin–1 ( x –  ) /( –  ) (C) 2 + +c (D) u +v + c
dx dx dx dx
(ii) 2sin–1 ( – x ) /( –  )
[B]
(iii) –2cos–1 ( – x ) /( –  )
x
51
Q.84 (tan –1 x  cot –1 x ) dx is equal to-
 1 
 x – 2 (  ) x 52
–1   (A) (tan–1 x + cot–1 x) + c
(iv) sin 52
1 
 2 ( – ) x 52
  (B) (tan–1 x – cot–1 x) + c
Then which of the above are correct – 52
(A) (i) only (B) (i) and (ii) only x 52 
(C) + +c
(C) (ii) & (iii) only (D) All [D] 52 2
x 52 
(D) + +c [A]
If  f ( x ) dx = f(x), then  [f ( x )]
2
Q.80 dx is 104 2
equal to
cos 4 x  1
Q.85  cot x – tan x dx equals-
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 1 1 sin x
(A) –
2
cos 4x + C (B) –
2
cos 4x + C Q.90  sin x – cos x dx equals-

1 (A) x + log (sin x – cos x) + c

(C) – cos 4x + C (D) None of these [C]
8 (B) 2 [x + log (sin x – cos x)] + c
cos 2 x  x  1 1
Q.86 x 2
 sin 2 x  2 x
dx equals- (C)
2
[x + log (sin x – cos x)] + c

(D) None of these [C]

(A) log (x2 + sin 2x + 2x) + c
dx
(B) – log (x2 + sin 2x + 2x) + c
Q.91  (x – 1) 3
( x  2) 5 
1/ 4 is equal to -

1 4  x – 1 1/ 4
(C) log (x2 + sin 2x + 2x) + c (A)   +c
2 3  x2
(D) None of these [C] 1  x  2 1/ 4
(B)   +c
3  x –2
x
55 x
Q.87 5 .5 5 .5 x dx is equal to-
1  x – 1 1/ 4
x
(C)   +c
55 3  x –2
(A) +c (B) 55
5x
(log 5)3 + c
(log 5) 3 1  x  2 1/ 4
(D)   +c
5x
3  x –1 
(C) 55 +c (D) None of these [C] [A]
(log 5)3
cos x – cos 3 x
Q.88 x 4
x
2
dx equals-
Q.92
 1 – cos3 x
dx equals-
 x 1
2
1  2x 2  1  (A) sin–1 (cos 3/2 x) + c
(A) tan –1   +c 3
3  3 

3
1  2x 2  1  (B) sin–1 (cos 3/2 x) + c
(B) tan–1   +c 2
3  3 
 
2
(C) cos–1 (cos3/2 x) + c
2  2x  1  2
3
(C) tan–1   +c
3  3  (D) None of these [C]
1  2x  1  2
(D) tan–1   +c
 [D]
3 3 x
  Q.93 x 4
–1
dx equals-
x
Q.89
 a – x3
3
dx is equal to –
(A)
1  x2 –1
log  2  +c
2 x 1 
3/ 2  
x
(A) sin–1   +c
a 1  x2 1
(B) log  2  +c

2 x
3/ 2 2  x –1
(B) sin–1   +c
3 a  x2 1
1
3/ 2 (C) log  2  +c

3 x 4  x –1
(C) sin–1   +c
2 a
1  x2 –1
3 x
2/3 (D) log  2  +c

(D) sin–1   +c 4  x 1 
2 a [D]
[B]
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  
dx  (A) –  x sin –1 x  1 – x2   c
Q.94 If f(x) =  sin 1/ 2
x cos 7 / 2 x
, then f   –f
4


(B) x sin–1x + 1 – x 2 + c


(0) =
x2
(A) 2.2 (B) 2.3 (C) +c
2
(C) 2.4 (D) 2.5 [C]
1
(D) (sin–1x)2 + c [A]
2
dt
Q.95 The value of t 2
 2 xt  1
(x2 > 1) is -
Q.100 Consider the following :
1  tx 
(A) tan –1  +c ax 2 – b
1– x2  1 x
2 
 If x c 2 x 2 – (ax 2  b) 2
dx =

1  t  x – x2 –1  ax  bx ax  b / x
  (i) sin–1 + k (ii) sin–1 +k
(B) log   +c
2 x2 –1 t  x  x 2
– 1 c c
 
ax 2  b ax 2  b
1 (iii) sin–1 +k (iv) sin–1 +k
(C) log (t2 + 2xt + 1) + c cx c
2
then which of the above are correct-
(D) None of these [B] (A) (ii) & (iii) only (B) (i) & (iii) only
(C) (ii) & (iv) only (D) (iii) & (iv) only [A]
dx
Q.96  (1  e x
)(1 – e – x )
equals -  3x – 4 
Q.101 If f   = x + 2, then
 3x  4 
 f ( x ) dx is
 ex –1   x 
(A) log  x  + c (B) log  e  1  + c equal to
  x 
 e 1   e –1 (A) ex +2 n
3x – 4
+c
3x  4
1  ex 1  1
(C) log  x +c
 (D) log 8 2
2  e –1 2 (B) – n |(1–x)|+ x+c
3 3
 ex – 1 8 x
  (C) n |(x – 1)|+ +c
 e x 1  +c [D]
3 3
 
(D) None of these [B]

dx
Q.97  2e x – 1
equals- Q.102
 
Let f : 0,   R be such that f(0) = 3 and
 2
(A) sec–1 2e x  c (B) sec–1  
2e x  c 1 
(C) 2sec –1
 2e x
c (D) 2sec –1 f '(x) =
1  cos x
.if a < f   <b, then a and
2
2e x
 c [D] b can be -

(A) , (B) 3 , 4
e x ( x  1) 2
Q.98  ( x  2) 2 dx is equal to-
(C) 3 +

,3 +

(D) 3 +

, 3 +
3
4 2 2 4
ex 2e x
(A) +c (B) +c [C]
( x  2) ( x  2)
ex 2e x e x ( 2 – x 2 )dx
(C)
( x  2) 2
+c (D)
( x  2) 2
+c [A] Q.103  (1 – x ) 1 – x2
-

ex
x n – x –n (A) +c (B) ex 1– x2 + c
Q.99 If f(x) = nlim

, 0 < x < 1, n  N then 1– x2
x n  x –n
 (sin
–1
x) f(x) dx is equal to -
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 e x (2 – x 2 ) e x (1  x ) 2  1 
3/ 2
  1  2
(C) + c (D) + c [D] (C) 1  2  log1  2    + c
1 – x2
2
1– x 3  x    x  3
(D) None of these [B]
x 4 1 2

–1
Q.104 If dx = tan f(x) – tan–1 g(x) + c,
6
x 1 3 Q.108
then -
1 sin 3 xdx
(A) f(x) = x +
x
, g(x) = x3  (cos 4
x  3 cos 2 x  1) tan –1 (sec x  cos x )
1 =
(B) f(x) = x2 + 2 , g (x) = x–3
x (A) tan–1(sec x + cos x) + c
1 (B) log tan–1 (sec x + cos x) + c
(C) f(x) = x2 – 2 , g (x) = x–3
x 1
1 (C) +c
(D) f(x) = x – , g(x) = x3 [D] (sec x  cos x ) 2
x
(D) None of these. [B]
10 x 9  10 x log 10
Q.105
 x 10  10 x
dx is equal to -
Q.109 The value of the integral
cos 3 x  cos 5 x
 sin 2
x  sin 4 x
(A) 10x – x10 + c
dx
(B) 10x + x10 + c
is - [IIT-1995]
(C) (10x – x 10)–1 + c
(D) log (10x + x10) + c [D] (A) sin x – 6 tan–1 (sin x) + c
(B) sin x – 2 (sin x)–1 + c
(C) sinx – 2 (sin x)–1 – 6 tan–1 (sin x) + c
1
Q.106 If  f (x) sin x cos x dx = log (D) sin x – 2 (sin x)–1 + 5 tan–1 (sin x) + c [C]
2( b 2 – a 2 )
1
(f(x)) + c, then f(x) is equal to -
1
Q.110  1  sin x
dx = ? [REE -2000]
(A)
a 2 sin 2 x  b 2 cos 2 x (A) 2 log tan(x/4 + /8)
1 (B) 2 log cot(x/2 + /8)
(B) 2 2 2 2
a sin x – b cos x (C) 2 log cos(x/2 + /8)
1 (D) None of these [A]
(C)
a 2 cos 2 x  b 2 sin 2 x
1
(D) [A] x2 1
a 2 cos 2 x – b 2 sin 2 x Q.111  x3 2 x 4  2x 2  1
dx is equal to–


x 2  1 log(x 2  1) – 2 log x  [IIT- 2006]
Q.107
 x4
dx is
(A) 2x 4  2x 2  1 + c
equal to - 2x 2
1  1 
1/ 2
  1  2 2x 4  2x 2  1 + c
(A) 1  2  log1  2    + c (B)
3  x    x  3 x3
1  1 
3/ 2
  1  2 2x 4  2x 2  1 + c
(B) – 1  2  log1  2  –  + (C)
3  x    x  3 x2
c 2x 4  2x 2  1 + c
(D) [A]
x

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 x {where ex 1 =
Q.112 Let f(x) = n 1/ n for n  2 and
(1  x ) t}
(fofo.........of ) = 2t–2 tan–1 t + C = 2 –1
g(x) =       (x).Then e x  1 –2 tan
f occurs n times
–1
( e x  1 ) + C = 2 fog (x) – 2 tan (fog (x)) +
x
n 2
g ( x ) dx equals-
[IIT-2007] C
1 1  A + B = 2 + (– 2) = 0
1
(A) n +K
n ( n  1) (1  nx ) n

dx
(B)
1
(1  nx n
)
1
1
n +K
Q.115  (x  1) x2 1
is equal to
n 1
1 1
1 x 1 x 1
(C) n +K (A) +c (B) 2 +c
n (n  1) (1  nx ) n x 1 x 1
1 1
1 x 1 x 1
(D) (1  nx n
) n +K [A] (C) +c (D) 2 +c
n 1 x 1 x 1
Sol.[C]
dx dx
Q.113 Let
 (x  1) 3/ 2
( x  1) 1/ 2
   x 1  1/ 2
p
  ( x  1) 2
e x  x 1
I =
 e4x  e2x  1 dx, J =
x 1
ut =t
x 1
e x
 e4x  e2x  1 dx Q.116  ( x  2) 7/8
dx
=
( x  3) 9 / 8
Then for an arbitrary constant C, the value of
1 1
J – I equals- [IIT-2008]
(A) 8  x  2  8  c (B) 5  x  2  8  c
1  e 4x  e 2x  1  5 x 3 8  x3
(A) log  4 x 2x
+C

2  e  e 1 1

(C) 5  x  3  8  c (D)
1  e2x  e x  1  8  x2
(B) log  2 x x
+C

2  e  e 1 
1

1  e2x  e x  1  8  x 3 8
(C) log  2 x x
+C


5 x2
 c
2  e  e  1 
Sol.[A]
1  e 4x  e2x  1 
(D) log  4 x 2x
+C
 [C]
2  e  e 1 1
Q.114 If f(x) = x , g(x) = ex –1, & fog (x) dx =A fog Q.117  tan 2   2
d( tan 2   1 ) is equal to
(x) + B tan–1(fog (x)) + C, then A + B is equal to
(A) 1 (B) 2 (A) n | tan 2   1  tan 2   2 |  c
(C) 3 (D) None of these (B) n | tan 2   1  tan 2   2 |  c

Sol.[D] fog (x) = ex 1 (C) n | tan 2   1  tan 2   1 |  c

(D) None of these
2t 2
 I=  ex 1 dx =
t 2
1
dt Sol.[A] Let tan2  – 1 = t2
dt
I= 
t2  3
= n | t + t2  3 | + c

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 (A) n | tan 2   1  tan 2   2 |  C
 ((e / x )
x x
Q.118  ( x / e) ) n x dx =
(B) n | tan 2   1  tan 2   2 |  C
x x x x
(A) (x/e) – (e/x) + c (B) (x/e) + (e/x) + c (C) n | tan 2   1  tan 2   1 |  C
x x x x
(C) (e/x) – (x/e) + c (D) (x/e) + 2 (e/x) + c (D) None of these
x Sol. [A]
x
Sol.[A] Let   =t
( x cos x  1)
 x 
e
x x 1
1
Q.122  2 x 3 e sin x  x 2
dx is equal to
   n x  x. x  .  dx = dt
 e  e e e 2 x  e sin x  1
  C
(A) n
x 
x
x  x
x
2 x  e sin x  1
   n  1 dx = dt    n x dx
e  e  e x e sin x  1  1
(B) n C
=dt x e sin x  1  1
 1  1
So I =   t 2
 1 dt =  + t + C
 t (C) n
2 x e sin x  1  1
C
x x 2 x e sin x  1  1
x e
=     C
e x 2 x e sin x  1  1
(D) 2n C
2 x e sin x  1  1
x 2  x –2 Sol. [C]
Q.119  x(x –2
–x ) 2 3/ 2
dx is –
Q.123  1  sin 2 x dx is
x x (A) sin x + cos x
(A) C (B) C
2 4
1– x 1– x (B) (sin x – cos x) sgn (sin x – cos x) + C
x (C) (sin x + cos x) sgn (cos x – sin x) + C
(C) C (D) None of these (D) (cos x + sin x) sgn (cos x + sin x) + C
1– x
Sol. [C]
Sol. [B]
 2x  3 
Q.124 If 3f   = 2x – 3, then
 2x  3 
 f ( x )dx is
( x cos x  1)dx
Q.120  2 x 3e sin x  x 2
is equal to equal to-
(A) ln(x – 1)2 + C (B) ln |(x – 1)3| + C
(C) ln (x – 1)6 + C (D) ln |(x – 1)| + C
2 x  e sin x – 1
(A) n C
2 x  e sin x  1 2x  3 3  1 t 
Sol. [A] Let =tx=  
2x  3 2  1 t 
x e sin x  1 – 1 2
(B) n C  f(t) =   f ( x )dx  ln(x – 1)2 +
x e sin x  1  1 t 1
C
2 x e sin x  1 – 1
e
tan 1 x
(C) n C Q.125 (1 + x + x2)d(cot–1 x) is equal to-
sin x
2x e 1 1
1 1
(A)  e tan x
C (B) e tan x
C
sin x
2x e 1 –1 1 1
(D) 2 n C (C)  x e tan x
 C (D) x e tan x
C
2 x e sin x  1  1
x
1 x  x 2 

1
e tan
Sol. [C] Sol. [C] I = –  1  x 2 dx
 
Let tan–1x = t
So I = –  e 1  tan t  tan 
t 2
1 t dt
Q.121  d ( tan 2   1) is equal to
 e  tan t  sec t dt
2
tan   2 =–
t 2

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 = – et . tan t + C = – x e tan x  C
1
(D) None of these

 x.x Let x2 = t 
2
sin x 1  sin x Sol.[B] g ( x 2 )dx
Q. 126 
sin 4 x
dx = A log
1  sin x 1
+ B log 2  tg( t )dt
1 2 sin x 1 1 2
C = [tF(t)–  F( t ) dt ] = [x F(x2)–
1 2 sin x
2 2
1 1
 F( x
2
) d( x 2 ) ]
(A) A = 8 , B = 4 2
dx
1
(B) A = – 8 , B = – 4 2
1 Q.130 The value of the integral  x n (1  x n )1 / n ;

1 1 nN is :
(C) A = – 8 , B = 4 2
1
1
1 1 (A) 1  1  n
(D) A = 8 , B = – 4 2 1 n 1  n 
  C
 x 
sin x sin dx
Sol.[C] sin 4 x
dx = 
2 sin 2 x cos 2 x (B)  1  1 
1  n 
1
1
n
C
1 n  x 
sin xdx
= 4 cos x sin x cos 2 x
=
1
1
(C) 1  1  n
1 cos x dx 1  n 
 C
 1 n 
 x 
4 cos 2 x cos 2 x
1
Q.127  x 3 (sin–1 (n x) + cos–1 (n x)) dx is
(D)  1  1 
1
n
1 n 1  n 
  C
 3/ 2  x 
(A) ( x  3) C (B)
3 Sol. [A]

(C) ( x  3) 3 / 2  C defined  ((e / x ) x
(D) NotQ.131  ( x / e) x ) n x dx :
2
Sol.[D] (A)(x/e) x – (e/x) x +C
Q.128 Let  e x {f ( x ) – f ' ( x )}dx = (x) then
(B) (x/e) x + (e/x) x +C

e
x
f (x) dx is (C) (e/x) x – (x/e) x +C

1 (D) (x/e) x + 2(e/x) x +C

(A) (x) + e f(x) x
(B) (x) + exf(x))
2 Sol. [A]
(C) (x) – e f(x) x
(D) e (x)
x
1
Q.132 
If I = e x tan–1 (ex) dx, then I equals:
e e
x x
Sol.[B] f ( x )dx – [exf(x) – f (x) dx] = (A) – e–x tan–1 (ex) + log (1 + e2x) + C
(x) 1
1 x (B) x – e–x tan –1 ex – 2 log (1 + ex) + C
e (e f(x) + (x)).
x
f ( x )dx =
2 1
(C) x – e–x tan–1 (ex) – 2 log (1 + e2x) + C
Q.129 Let  g ( x ) dx = F(x), then
(D) None of these
Sol.[C] put ex = t to get
x
3
g ( x 2 )dx equals to -
1
1 2 I=  t2 tan–1 (t) dt

2
(A) [x (F(x))2 – ( F( x )) dx]
2 1 1  1
1 2 2 = (tan–1 t) t –  1  t 2   t  dt
 (F( x
2
(B) [x F(x ) – )) d (x2)] 1 1
2
= – t tan–1 t +  t (1  t 2 ) dt
1 2 1
 (F( x ))
2
(C) [x F(x) – dx]
2 2
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 1 1 t  a 2 x  b 2 x  2a x b x
 
= – t tan–1 t +  t 1  t 2 
 dt =  a xbx
dx
1 1
= – t tan–1 t + nt – 2 n (1+ t2) + C  ax bx 
=   b x  a x  2  dx
Q.133 If xf(x) = 3f 2(x) + 2, then  

2 x 2  12xf ( x )  f ( x ) 
 a 
x
b
x 

 (6f (x )  x )(x 2  f ( x )) 2 dx equals- =   b      2 dx
a
  
 
1 1
(A) +c (B) +c (a / b ) x (b / a ) x
x 2  f (x) x 2  f (x) = + + 2x + c.
log e (a / b) log e ( b / a )
1 1
(C) +c (D) +c Hence (D) is the correct answer.
x  f (x) x  f (x)
f (x)
Sol.[A] f  (x) =
6f ( x )  x
sin n x
2 x ( x  6f ( x ))  f ( x ) Q.136  3
cos n  6 x
dx, n  N is equal to -
Now I =  (6f (x )  x )(x 2  f ( x )) 2 dx
n
2 x  f ( x ) 1
1 (A) 3 (tan x ) 3  c (B)
I=–  ( x 2  f (x )) 2 dx =
x 2  f (x)
+C n

n
e x  e  x  (e x  e  x ) sin x 3 1
Q.134
 1  cos x
dx =
3 n
(tan x ) 3  c

(A) (ex + e–x) tan (x/2) + C 3

x –x (C) (cos x ) n 1  c (D) None of
(B) (e – e ) cot (x/2) + C n
(C) (ex – e–x) sec (x/2) + C these
(D) (ex – e–x) cosec (x/2) + C
sin n x sin n x
Sol. [A] Sol.[B]  3 dx = 3 cos n x . cos 6 x
dx
(a x  b x ) 2 cos n  6 x
Q.135  a xbx
dx is equal to –
3
=  tan n x sec 2 x dx

(b / a ) x (a / b ) x
(A) + +x+c
 (tan x )
n/3
log e (a / b) log e ( b / a ) = d (tan x)

n
(a / b ) x (b / a ) x 1
(B) + +x+c (tan x ) 3
log e ( b / a ) log e (a / b) = +c
n
1
(a / b ) x (b / a ) x 3
(C) + + 2x + c
log e ( b / a ) log e (a / b)
n
(a / b ) x (b / a ) x = 3 1
(D) + + 2x + c (tan x ) 3 + c.
log e (a / b) log e ( b / a ) 3 n

(a x  b x ) 2
Sol.[D]  a xbx
dx Hence (B) is the correct answer.

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 1 1 x 1
Q.137  3 dx is equal to - We get I =  dx
sin x sin( x  a ) 1 x x

2 1  cos 2 1
(A) –
sin a
cos a  cot x sin a  c =  1  cos 2
×
cos 2
× (–2 sin 2)

2 d
(B) cos a  cot x sin a  c
sin a
2 sin 2    4 sin  cos  
(C) –
1
cos a  cot x sin a  c
=  2 cos 2 
× 
 cos 2
 d

sin a
4 sin 2 
(D)
1
cos a  cot x sin a  c
=–  cos 2
d
sin a
1  cos 2
Sol.[A] We have 
1
dx
= –2  cos 2
d
sin 3 x sin( x  a )
= –2 (sec 2 – 1) d
1
=  dx
sin 3 x (sin x cos a  cos x sin a ) = –2 log |sec 2 + tan 2| + 2 + c
1 1  sin 2
=  4 dx = –2 log
cos 2
+ 2 + c
sin x (cos a  cot x sin a )

1  sin 2
cos ec 2 x = –2 log + 2 + c
=  cos a  cot x sin a
dx 1  sin 2 2

1 1 1  sin 2
=–
sin a
 cos a  cot x sin a
d (cos a + = –2 log
1  sin 2
+ 2 + c

cot x sin a) 1 1 x2
= – log 2
+ cos–1 x + c
1 1 1 1 x
=–
sin a
 t
dt, where t = cos a + cot x
Hence (C) is the correct answer.
sin a 1
2
Q.139  (sin 2 x  cos 2 x ) dx = 2
sin (2x – a) +
=– cos a  cot x sin a  c .
sin a b
Hence (A) is the correct answer.
5 5
(A) a = , b  R (B) a = – ,bR
1 x 1 4 4
Q.138  1 x
·
x
dx is equal to -

(C) a = ,bR (D) None of these
1 x  1 x 4
(A) –2 log + cos–1 x + c
1 x  1 x
1
1 x  1 x Sol.[B]  (sin 2x – cos 2x) dx = sin (2x – a) + b
(B) – log –1
+ cos x + c 2
1 x  1 x
 1 1 
(C) – log
1 1 x2 –1
+ cos x + c
 2   2
sin 2 x 
2
cos 2 x  dx

1 1 x2
1
(D) None of these = sin (2x – a) + b
2
Sol.[C] On putting x = cos 2, and dx = –2 sin 2 d,

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  1 1   3y  2 
– 2   2
cos 2 x 
2
sin 2 x  dx

x3 = ln 
 3 


1  3y  2 
= sin (2x – a) + b or x= 3 ln .
2  3 
1
– 2  cos (2x + /4) dx = 2
sin (2x –
cos 4 x  1
a) + b
Q.141  cot x  tan x dx is equal to -
2 1 (A) –1/2 cos 4x + c (B) –1/4 cos 4x + c
– sin (2x + /4) + c = sin (2x – a) +
2 2 (C) –1/2 sin 2x + c (D) None of these
b
 2 sin 2 2 x
1  5  1 Sol.[D] 
cos 4 x  1
dx =  cos 2 x dx
 sin  2 x   +c= sin (2x – cot x  tan x
2  4  2 sin x cos x
a) + b sin 2 x (1  cos 2 2 x )
5
=–  cos 2 x
dx
a=– , b  R.
4 dt
Put cos 2x = t  – sin 2x dx =
2
Q.140 Let the equation of a curve passing through the 1 1 t 2 1 t2
3
=
2  t
dt =
2
log t –
4
x
2
point (0, 1) be given by y = .e x dx. If
+c
the equation of the curve is written in the form x
1 cos 2 2 x
= ƒ(y) then ƒ(y) is - = log cos 2x – + c.
2 4
 3y  2   2  3y 
(A) ln 
 3 
 (B) 3 ln 
 3 
 Q.142 If  cosec 2x dx = ƒ (g(x)) + c, then -

 3y  2  (A) range g(x) = (–, )

(C) 3 ln  (D) None of these
 3  (B) dom ƒ(x) = (–, )
3 (C) g (x) = sec2 x
x
2
Sol.[C]  y= .e x dx
(D) ƒ(x) = 1/x for all x  (0, )
1 x3
 3x
2
= .e dx
3 Sol.[A]  cosec 2x dx = ƒ (g(x)) + c

1 3
y= . ex + c
3  cosec 2x = ƒ(g(x)) g(x)
it is passing through (0, 1) then 1
 × sec2 x = ƒ (g(x)) g(x)
1 2 2 tan x
1= +cc=
3 3 1
  ƒ(x) = , g(x) = sec2 x
1 x3 2 2x
Then y = e +
3 3 Dom. ƒ(x) = (–, ) – {0}, g (x) = tan x + c
3 (3y  2) Range g(x) = (–, ).
 ex =
3

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Q.143 If 
xe x
dx = ƒ(x) (1  e x ) – 2 log
Q.144  x sin 2x . cos 3x dx is -
x
(1  e )
x cos 5x sin 5x
g(x) + C, then - (A) – + +c
10 50
(A) ƒ(x) = x – 1 (B) g (x) =
x cos 5x sin 5x x cos x
(B) – + + –
10 50 2
1 ex 1
1 ex 1 sin x
+c
1 ex 1
2
(C) g(x) = (D) ƒ(x) =
1 ex 1 x cos 5x sin 5x x cos x
(C) – + – +
10 50 2
2(x – 2)
xe x sin x
Sol.[B] Let I =  x
dx
2
+c
(1  e )
(D) None of these
ex
=  x · dx
Sol.[B] I = x sin 2x . cos 3x dx =
(1  e x )

I II
1 1
=x.2 (1  e x ) –  1.2 (1  e x ) dx
=
2
 2x cos 3x . sin 2x dx =
2
x (sin 5x –

sin x) dx
= 2x (1  e )x
–2 (1  e x ) dx
1 1
in second integral
=
2
x sin 5x dx–
2
x sin x dx

put 1 + e x = t2 …(1)

 dx =
2 t dt
2 Let I1 =  xI sin
5
 cos 5x 
 x dx = x  
5
 –
t 1 II  

then = 2x (1  e x ) – 4 1 .

t 2 11  cos 5x 
 (t 2
 1)
dt 
 5 
 dx

 1  x cos 5x sin 5x
= 2x (1  e x ) –4 1   dt =– +
2 5 25
 t 1 


& I2 =  xI sin
 x dx = x(–cos x) –  1 . (–
 1  t  1  II
= 2x (1  e ) x
– 4 t  log  + c

 2  t  1  cos x)

 1  ex 1  dx = –x cos x + sin x
 
= 2x 1  e x –4 (1  e ) – 2 log 
x +c Put I1 & I2 in equation (1)
 1  e x  1 
 
On comparing

1 ex 1
ƒ(x) = 2x – 4, g(x) = .
1 ex 1

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  I = –
x cos 5x sin 5x x cos x
+ + –
1 x 11  1
(B) log +c
10 50 2 2 x 11  1

sin x (C) 2 1  X11 + c

+c
2 (D) None of these
11 9/2
Sol.[D] Let x11/2 = t  x dx = dt
e ax 2
e
ax
Q.145 .sin bx dx is (a sin bx – b cos bx)
l
2 dt 2
+ c. Then l is -  I =  2 = log
11 1 t 11
(A) a2 + b2 (B) (a 2  b 2 ) t  1 t2 




+c
2 2 2
(C) (a + b ) (D) None of these
2  11 / 2 
1  x11 
= log x  + c.
e ax

ax sin bx  
Sol.[A] I = e .    dx = sin bx . – 11
II I a
 dx 
Q.147   sin x  sec x  is -
e ax
 b cos bx . a
dx
1 tan x  3
(A) ln + tan–1 (sin x + cos
2 3 tan x  3
ax b
e . sin bx
e
ax
= – . cos bx dx
a a x) + c
1 sin x  cos x  3
e ax . sin bx b (B) ln + tan–1
= – 2 3 sin x  cos x  3
a a
(sin x + cos x) + c
 e ax e ax 
 . cos bx   (  b sin bx )dx 
 a a  1 tan x  3
(C) ln + tan–1 (sin x – cos
2 3 tan x  3
e ax . sin bx b . e ax b2
= – . cos bx – x) + c
a a2 a2
(D) None of these
dx cos x dx
 e  sin
ax
bx dx
  
Sol.[B] I =  (sin x  sec x ) =  1  sin x cos x
I

 b2  e ax 1 cos x  sin x 1
 I 1  2  = 2 (a sin bx – b cos bx) =  1  sin x cos x dx +
 a  a 2 2

e ax cos x  sin x
I= 2
(a  b ) 2 (a sin bx – b cos bx) + c
 1  sin x cos x dx
Hence l = (a2 + b2). Let sin x – cos x = t Let sin x + cos x = u

x9 / 2  (cos x + sin x) dx = dt  (cos x – sin x) dx = du

Q.146  dx is -
1  x11

2  x7 / 2  x7 1 
(A) log 



+c
11

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 e
dt t
+
So I =
1  1 t2  + 1
2 1  
 2  2
  1
dt
t2

du II I II I
  u 2 1  1 1
 e
t
1   = (log t) e t – . et dt + . dt +
 2  t t
 

dt du 1 t 3
=  3  t2 +  u2 1 = 2 3
ln
t 3
 1 t
  e
 t
+ tan–1 u + c
1 1
 t2 e
t
– . et dt + dt +
1 sin x  cos x  3 –1 t2
= ln +tan (sin x +
2 3 sin x  cos x  3
c
cos x) + c
1 t
= et . log t – e +c
 (B) is correct t

 1 
 x
Q.148  log(log x )  (log x ) 2  dx is equal to - = x log (log x) –
log x
+ c.
 
Hence (A) is the correct answer
x
(A) x log (log x) – +c 1 1
log x Q.149  cot (1 – x + x2) dx = x tan–1 x –
2
log (1 +

x
(B) x log x – +c x2) + A, Then A is equal to -
log x
1
(A) – (1 – x) tan–1 (1 – x2) + log {1 + (1 – x)2} +
x 2
(C) xx log x – +c
log x c

x (log x ) 1
(D) x log x – +c (B) (1 – x) tan–1 (1 – x2) + log {1 + (1 – x)2} +
log x 2
c

 1 

Sol.[A] Let I =  log(log x )  2  dx 1

 (log x )  (C) – (1 – x) tan–1 (1 – x) + log {1 + (1 – x)2} +
2
 1
 I=  log t  t 2  e dt, where t = log x
t c
(D) None of these
 1 1 1 t

 log t     e dt
t t t2  Sol.[C] We have,  cot
1
(1 – x + x2) dx

 1  1 1 1
  log t  t  e t
dt +  – t  t 2  e t =  cot {1 – x (1 – x)} dx

dt 1  1 
=  tan   dx
1 1  x (1  x ) 
  e log t dt + e e
t t t
. dt +
t
(–1/t) dt
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  x  (1  x )  Hence (C) is the correct answer.
=  tan 1   dx
1  x (1  x ) 
x 3  3x  2 1
=  {tan
1 –1
x + tan (1 – x)} dx
Q.150  (x 2  1) 2 (x  1) dx = 4
log (x2 + 1) +

1 1
=  tan x dx +  tan (1 – x
+
3
tan–1 x – A. Then A is equal to -
2
x) dx x 1 2
3 2
(A) log |x + 1| + c (B) log |x + 1| + c
= I1 + I2 … (1) 2 3
1 1
1 1 (C) log |x + 1| + c (D) log |x2 + 1| + c
where I1 =  tan x dx and I2 =  tan (1 – 2 2
x) dx. x 3  3x  2
1
Sol.[C] Let I =  (x 2  1) 2 (x  1) dx, then
Now, I1 =  tan x dx
1 ( x 3  x )  2( x  1)
=  tan x 1 dx I=  dx
( x 2  1) 2 ( x  1)
I II x 1
=  ( x 2  1)(x  1) dx + 2  ( x 2  1) 2
x
= x tan–1 x –  1 x2 dx dx
= I1 + 2I2, where
1 1
= x tan x – –1
 1 x2 d (1 + x
2 I1 =  ( x 2  1)( x  1) dx and I2 =
x2)
1 1
= x tan–1 x – log (1 + x2) … (2)
2  ( x 2  1) 2 dx
and
x
I2 =  tan 1 (1 – x) dx Now, I1 =  ( x 2  1)( x  1) dx
1
= –  tan (1 – x) d (1 – x) 1  x 1 1 
= –
= 2    dx
 x 1 x 1
2
[Using partial

fractions]
 1 1 2 
(1  x ) tan (1  x )  2 log{1  (1  x ) } 1 x 1 1 1
  =
2  x 2  1 dx + 2  x 2  1 dx – 2
[using (2)]
1
Substituting the values of I1 and I2 in (1), we get  x  1 dx
1 1
 cot (1 – x + x2) dx = x tan–1 x –
2
log (1
=
1
log (x2 + 1) +
1
tan–1 x –
1
log |x + 1|
4 2 2
+ x2)
+ c1
1
– (1 – x) tan–1 (1 – x) + log {1 + (1 – x)2} + c
2
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 1 1
and, I2 =  ( x 2  1) 2 dx
=
1  1
2  1
dx  
2 x    2  x
 x
1 x 1
= 2 + tan–1 x + c2
2 ( x  1) 2 1
–
1  1
2  1
d x  
Hence, I = I1 + 2I2 2 x    2  x
 x
1 x
= log (x2 + 1) + 2
4 x 1 1 du 1 dv
=
2
 u2  ( 2) 2 –
2
 v2  ( 2 )2
,

3 1
+ tan–1 x – log |x + 1| + c. 1 1
2 2 Where u = x – and v = x +
x x
Hence (C) is the correct answer.
1 1  u  1
= × tan–1   – 1 × log
 x2 1 2 2  2 2 2 2
1 1
Q.151 If  x4 1 dx =
2 2
tan–1 
 2 x
 +A


v 2
+ c. Then A is equal to - +c
v 2
1 x2  2x  1
(A) – log 1  x 2 1  1
2 2 x 2
 2x  1 = tan–1   – log
2 2  2x  4 2
 
1 x2  2x  1
(B) – log
4 2 x2  2x  1
x2  2x  1
2 +c
1 x2  2x  1 x  2x  1
(C) – log
2 x2 – 2x  1 Hence (B) is the correct answer.

1 x2  2x  1
(D) – log 2
 sec
4 2 4/3
x – 2x  1 Q.152 x cosec8/3 x dx is equal to -
1 3
Sol.[B]  x 4  1 dx (A) –
5
tan–8/3 x + 3 tan–2/3 x + c

1/ x 2 3
=  x 2  1/ x 2 dx (B) –
5
tan–5/3 x + 3 tan1/3 x + c

3
1 2 / x2 (C) tan–8/3 x + 3 tan–2/3 x + c
=
2  x 2  1/ x 2 dx 5
3
 1   1  (D) – tan–8/3 x – 3 tan–2/3 x + c
1  2   1  2  5
1  x   x 
=
2  1
dx Sol.[B] We have, I =  sec
4/3
x cosec8/3 x dx
x2  2
x 1
 cos 4 / 3 x sin 8 / 3 x dx =  cos
4 / 3 –
= x sin
1 1
1 1–
x 2 dx – 1
8/3
1 x2 x dx
= 
2 2
dx
2 x2  1 x 
1
x2 x2

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 4 8 of ƒ(x) does not have an extremum at x = 0, then
Since –    = –4, which is an even
 3 3
ƒ( x )
integer. So, we divide both numerator and the value of the integral  x 3 1 dx is -
denominator by cos4 x.
x2
sec 4 x (A) c (B) x + c
I=  tan 8 / 3 x dx 2
x3
(1  tan x ) 2 (C) c (D) None of
=  tan 8 / 3 x
sec2 x dx b
these
2
1 t Sol.[B] Let ƒ(x) = ax 3 + bx2 + cx + d. Then,
=  t 8/3
dt where t = tan x
ƒ(0) = –1 and ƒ(1) = 0
 d = –1 and a + b + c + d = 0
3
=  (t
–8 / 3
t 2 / 3
) dt = – t – 5 / 3  3t 1 / 3 +
5  d = –1 and a + b + c = 1 … (1)

c It is given that x = 0 is a stationary point of ƒ(x)

but it is not a point of extremum. Therefore,
3
=– tan 5 / 3 x  3 tan1/3 x + c.
5 ƒ(0) = 0, ƒ(0) = 0 and ƒ(0)  0

Hence (B) is the correct answer. Now ƒ(x) = ax3 + bx2 + cx + d

 ƒ(x) = 3ax2 + 2bx + c, ƒ(x) = 6ax + b and ƒ(x) =

77
x
x 6a
Q.153 7 . 77 . 7x dx is equal to -
 ƒ(0) = 0, ƒ(0) = 0 and ƒ(0)  0
x x
77 77
(A)
7
+c (B)
7
+c  c = 0, b = 0 and a  0 …(2)
3
(log e 7) (log e 7) 2 From equations (1) and (2), we get
7x a = 1, b = c = 0 and d = –1.
(C) 7 7 . (loge 7)3 + c (D) None of these
7x x  ƒ(x) = x3 – 1
Sol.[A] We have, 7 . 77 . 7x dx
7

ƒ( x )
1  77 x   7 7 log 7 
x
Hence,  x 3 1 dx =  1 dx = x + c.
=   7 log e 7 
 .  e 
(log e 7) 3   
Hence (B) is the correct answer.
.

7 x
log e 7 dx  Q.155 If  ƒ( x ) dx = ƒ(x), then  {ƒ( x )}
2
dx is
7x equal to -
1 77
3  1 . d( 7
= 7x ) = +
(log e 7) (log e 7) 3 1
(A) {ƒ(x)}2 (B) {ƒ(x)}3
2
c.
Hence (A) is the correct answer. {ƒ( x )}3
(C) (D) {ƒ(x)}2
3

Q.154 Let ƒ(x) be a polynomial of degree three satisfying Sol.[A] We have,  ƒ( x ) dx = ƒ(x),
ƒ(0) = –1 and ƒ(1) = 0. Also, 0 is a stationary point d
 (ƒ(x)) = ƒ(x)
dx

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 1 dt

ƒ( x )
d (ƒ(x)) = dx = t t 1

 log (ƒ(x)) = x + log c ydy

 ƒ(x) = cex
=2  ( y 2  1) y (where y2 = t – 1)

{ƒ(x)}2 = c2e2x = 2 tan–1 y + C = 2 tan–1 sec x – 1 + C

c 2e 2x  ƒ(x) = tan–1 x and g(x) = sec x – 1 .

  {ƒ( x )} dx = c
2 2
e 2 x dx = =
2 dx
Q.158 If  =  (goƒ) (x) + C, then -
x 5x 2  3
1
{ƒ(x)}2.
2 5 2 1
(A) g(x) = tan–1x, ƒ(x) = x 1 ,  =
Hence (A) is the correct answer. 3 3

Q.156  x | x | dx = (B) g(x) =

5 2
x  1 , ƒ(x) = tan–1 x,  =
3
x3 x2 | x |
(A) (B)
3 3 1
2 3
x |x|
(C) (D) None of
2 1
(C) g(x) = tan–1 x, ƒ(x) = 5x 2  3 ,=
these 2
Sol.[B] I = x . |x| dx = | x | . x dx
1
(Integrating by parts taking |x| as first function) 5
(D) None of these
x2 | x | x2 x2 | x |
= |x| .
2
–  x . 2 dx = 2 Sol.[A] Put 5x2 – 3 = t 2  5x dx = t dt
x 1 5t dt
–
1
 | x | x dx I=  2 2 dx =
5
 ( t 2  3) t
2 x 5x 3

 1 x2 | x | dt
 I 1   =
 2
1 2
I= x |x|.
3
=  t2  3
2

1 t 1
= tan–1 + C = tan–1
Q.157 If  1  sec x dx = 2 (ƒog) (x) + C, then - 3 3 3

(A) ƒ(x) = sec x – 1 (B) ƒ(x) = 2 tan–1x

(C) ƒ(x) = sec x  1 (D) None of 5x 2  3
+ C.
3
these
sec x. tan x 5 2
 g(x) = tan–1 x and ƒ(x) = x 1 .
Sol.[C]  1  sec x dx =  sec x. sec x 1
dx 3

tan x x2
=  sec x – 1
dx Q.159 If ƒ(x) =
1 x2
and g(x) = sin x, then

(Put sec x = t  sec x tan x dx = dt)

 (ƒog ) (x) cos (x) dx =

(A) sin x – tan–1 (sin x) + c

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 (B) cos x – tan–1 (sin x) + c (C) cos x – sin x + c (D) None of these
(C) cos x + tan–1 (sin x) + c Sol.[B] Do your self.

(D) sin x – tan–1 (cos x) + c Q.162  cos 2 x sin4x dx is equal to -

1
Sol.[A] I =  (ƒog ) (x) cos (x) dx = (A) (cos 6x + 3 cos 2x ) + c
12
 (ƒ(g ( x ))) cos x dx 1
(B) (cos 6x + 3 cos 2x) + c
6
sin 2 x 1
=  ƒ(sin x ) cos x dx =  1  sin 2 x cos x (C) –
12
(cos 6x + 3 cos 2x) + c

dx. (D) None of these

Sol.[C] Do your self.
t2
=  1  t 2 dt (where sin x = t  cos x dx =
sin x
dt)
Q.163  1  cos x
dx equals -

1 (A) 2 cos(x/2) + c
= 1 – dt = t – tan–1 t + c
1 t 2 (B) 2 sin(x/2) + c

= sin t – tan–1 (sin t) + c. (C) 2 2 cos(x/2) + c

(D) – 2 2 cos(x/2) + c
Sol.[D] Do your self.
Q.160 If l r means log log log ….. x the log being r

 tan
3
Q.164 x sec 2 x dx-
times then  [x l(x) l2 (x) l3 l4 (x)…. l r (x)]–1 dx =
1
 (tan x )
3
r+1 l r 1 ( x ) (A) tan4 x + c
d (tan x) =
4
(A) l (x) (B)
r 1 1
(B)  (cos x ) d (tan x) =
3
tan4 x + c
(C) lr (x) (D) None of these 3
1
1 (C)  (tan x ) d (tan x) = –
3
tan4 x + c
Sol.[A] I1 =  x log x dx 4
(D) None of these
1 Sol.[A] Do your self.
Put log x = t  dx = dt
x
(log x )
 I1 = 
1
dt  log t = log (log x) = l2 (x)
Q.165  x2
dx =
t
1 1
(A) (log x + 1) + c (B) – (log x + 1) + c
1 2 x
I2 =  x (log x ) log(log x ) dx 1
(C) (log x – 1) + c (D) log (x + 1) + c
Put log (log x) = l2x = t x
Sol.[B] Do your self.
1 1 {f ( x ).'( x ) – f '( x )( x )}
 . dx = dt Q.166  {log(x)–
log x x f ( x ).( x )
logf(x)}dx =
1
 I2 =  t dt = log t = log log (log x) = l3 x (A) log
( x )
+K
f (x )
 Ir = lr + 1 (x)  (A). 2
1  ( x ) 
(B) log  +K
Q.161  1  sin 2 x dx equals- 2  f (x) 
(A) sin x + cos x + c (B) sin x – cos x + c
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 ( x ) ( x )  x sec x
(C) .log +K (B) + tan x + c
f (x ) f (x ) x sin x  cos x
(D) None of these  x sec x
(C) + sec x tan x + c
( x )  ( x )  x sin x  cos x
Sol.[B] I = log d log
f (x)  f (x) 

 x sec x
(D) + 1 – tan2 x + c
1  ( x ) 
2 x sin x  cos x
= log  +K
2  f (x)  [B]

Q.167 e
x
{f ( x ) – f ' ( x )} dx = (x) then (1  x ) 2
Q.135  ex
(1  x ) 2
dx is equal to
e
x
.f ( x ) dx =
1 ex
(A) (x) + ex . f(x) (B) (x) – ex . f(x) (A) 2
c (B) c
x 1 x2 1
1 1
(C) {(x) + exf(x)} (D) {(x) + ex f '(x)} ex 1 1 ex
2 2 (C) c (D) c [B]
x2 1 x2 1
e (f(x) – f '(x)) dx = (x)
x
Sol.[C] Here

e
x
& (f(x) – f '(x)) dx = ex.f(x) dx
Q.136  x(x n
 1)
is equal to :
 .e
x
Add  2 . f(x)dx = (x) + ex f(x)
1  xn 
(A) log n +C
 (B)
dx n  x 1 
Q.168  ( x  1)( x – 2)
= A log (x + 1) + Blog(x–
1  xn 1
2) +c then - log n
 +C

n  x 
(A) A + B = 0 ; A : B = – 1
(B) A B = 1 ; A : B = – 1  xn 1
(C) log n
 +C (D) None of these
 [A]
(C) A – B = 0 ; A : B = – 1  x 
(D) None of these e x (1  sin x )
Sol.[A] I =
1 ( x  1) – ( x – 2)
 dx
Q.137
 1  cos x
dx is equal to –
3 ( x  1)( x – 2)
1
1  1 1  (A) log tan x (B) ex tan x
= 
 –
3  ( x – 2) x 1 
 dx
(C) sin log x
2
(D) ex cot x. [B]
1 1
= log(x – 2) – log(x + 1) + c Q.138 If x sin x dx = – x cos x + , then  =
3 3
1 (A) sin x + c (B) cos x + c
B= & A = – 1/3 (C) x cos x + c (D) cos x – sin x + c. [A]
3

e ( x 4  1) 1 dx 
3 log x
x  sin x Q.139
Q.133  1  cos x
dx is equal to
(A) log (x4 + 1) + c (B) 3 log(x4 + 1) + c
x x
(A) – x cot +c (B) cot +c 1
2 2 (C) – log(x4 + 1) + c (D) log(x 4  1)  c
4
x
(C) – cot +c (D) None of these [A] [D]
2
x2 x
Q.134  (x sin x  cos x ) 2
dx is equal to Q.140  a  x3
3
dx  g ( x )  c , then g(x) =

x sec x
(A) + tan x + c
x sin x  cos x
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 2 2  x3   x7  1  x7 
(A) cos –1 x (B) sin  3  (A) log 7  + C(B)
 log 7  +

a 
3 3    x 1  7  x 1 
2 C
(C) sin –1 ( x 3 / a 3 ) (D)
3  x7 1  7 
2 (C) log  + C(D) 1 log x  1  +
cos –1 ( x / a ) 7  7  x7 
[C]  x   
3
C
sin x [B]
Q.141 If  sin( x  ) dx  Ax + B log sin(x – ) dx
+ c, then value of (A, B) is –
Q.145 x 2
 4x  13
is equal to -
(A) (– sin , cos ) (B) (cos , sin )
(A) log(x2 + 4x + 13) + C
(C) (sin ,cos ) (D) (– cos , sin ) [B]
1  x2
2 (B) tan –1   +C
x2 3  3 
Q.142   x  4  ex dx is equal to -
(C) log(2x + 4) + C
x x  x x  2 2x  4
(A) e  c (B) e  c (D) +C [B]
x4 x4 ( x  4 x  13) 2
2

x x – 2  x
2 xe  
(C) e  c (D)  
c Q.146  cos
–3 / 7
x sin–11/7 x dx is equal to -
 x4  x4
4
xx  4x  4 
2 (A) log | sin4/7 x | + C (B) tan4/7 x + C
Sol.[A]  e  2 
 ( x  4) 
dx
 =
7
7
 x 4  (C) – tan–4/7 x + C (D) log | cos3/7 x | + C [C]
 ex  
 x  4 ( x  4) 
2  dx 4
1 x  x  x2

F( x )

F( x )
Q.147  x  1 x
dx is equal to -

= ex (x/x+ 4) + C 1 2
(A) 1 x + C (B) (1 + x)3/2 + C
2 3
(sin   cos )
Q.143  sin 2
d is equal to - (C) 1 x + C (D) 2(1 + x)3/2 + C

1 x  x  x2
(A) log | cos  – sin  + sin 2 |  c Sol.[B] I=  x  1 x
dx=

(B) log | sin  – cos  + sin 2 |  c

(C) sin–1 (sin  – cos ) + c (1  x ) 2  x 1 x
(D) sin–1 (sin + cos ) + c
  x  1 x  dx

Sol.[A] In =  tan
n–2
x tan 2 x dx
1 x  x x 1  2
=  tan
n –2
x (sec 2 x – 1)dx In
=   x x 1  dx =
3
(1 + x)3/2

=  tan
n –2
x sec 2 xdx – I n – 2 +C
tan n –1 x
tan x = t 2 I
sec xdx = dt n  – In –2
n –1

Put n = 6I6 + I4 = tan5x/5 Both state. are correct
and correct expla.

dx
Q.144  x(x 7
 1)
is equal to -

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Q.148 dx
Q.150  (sin x – 2 cos x)(2 sin x  cos x ) is
dx
 x (log x )(log log x )...(log log ...x )
    
equal to -

8 times tan x – 2
(A) log e 5 +C
is equal to - 2 tan x  1

(log log ...x )  C 1  tan x – 2 

(A)      (B) (B) log  +C
8 times 4  tan x  2 

(log log ...x )  C

1  2 sin x  cos x 
     (C) – log  +C
7 times
4  sin x  2 cos x 
(D) None of these [A]
(log log ...x )  C
(C)      (D) None of
9 times

these
Sol.[C]

dx Q.151 If f(0) = f '(0) = 0 and f ''(x) = tan2x, then f(x) is

 x (log x )(log log x )...(log log ...x )
     x2 x2
8 times
(A) log sec x – (B) log cos x +
2 2
log log ...x
put      = t x2
8 times (C) log sec x + 2 (D) None of these [A]
I =
1 x 24
 x10  1 dx ; where t = x .
5
Q.152 The value of
(log log ...x ).....(log log x )(log x ) x
    
7 times
1  t3 1 
= dt (A)   t  tan t  (B)
5  3 
dt (log log ...x )
I=  t
= log t + C = log     
8 times  t3


 t  tan 1 t 
 3 
+C  
x2 –1 1  t3 
Q.149  (x 4  1  dx is equal
(C)
5

 3

 t  tan 1 t  (D)


 3x 2  1) tan –1  x  
 x
 t3 
to -   t  tan 1 t 
 3 
 
–1  1
(A) tan  x    C x 24 x 20 .x 4
 x Sol.[A]  x 10  1 dx =  ( x 5 ) 2  1 dt Put x 5
=t

–1  1
(B) cot  x    C 1 t4 1 t 4 1  1
 x 
5  t 2 1 dt =
5  t2 1
dt
 1
(C) log x    C 1  1  1
  t
2
 x = 1  dt =
2
5 1 t  5
 –1  1 
(D) log  tan  x     C [D]  t3
  x   1 

 3  t  tan t  , t = x
5

 

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 cos 4 x  1  2 1 1
Q.153 If
 cot x  tan x
dx = a cos22x + b cos 2x  2 x log x  x .  dx =
 x t
dt

+  t (2 log x + 1) x dx = dt
x
c log | cos 2x | +  being a constant of  ( x x ) (2 log x + 1) x dx = dt
integration then -
1
so question becomes =  1.dt =t+c
(A) a = – (B) b = –
8 sin x
1 Q.155 If  sin(x  ) dx = Ax + B log sin (x – )
8 + c then value of (A, B) is
5 (A) (– sin , cos ) (B) (cos , sin )
(C) c = (D) a + b + c = 0 (C) (sin , cos ) (D) (– cos , sin )
8
sin( x     )


cos 4 x  1
1
Sol.[B]  sin( x   )
dx
Sol.[B] cos x sin x dx =
 2 sin(x   ) cos   cos(x   ) sin 
sin x cos x =  sin(x   )
(cos 4 x  1) sin 2 x
 cos 2 x
dx dx
= cos   dx  sin   cot( x  )dx

 1  cos 2 x 
2 

1    1 = (cos ). x + sin . log {sin (x – )} + c
= 2 sin 2x dx
2    

 cos 2 x
1  (1  cos 2 x ) 2  4 
  
 1 

=
8  

 cos 2 x
 sin 2x dx


Q.156  log(log x )  (log x ) 2  dx =


put cos 2x = t  (– sin 2x) 2dx = dt (A) log (log x) + c
dt (B) x log (log x) + c
sin 2x dx = –
2  1 
(C) x log(log x )   +c
 log x 
1 (1  t ) 2  4
=–
16  t
dt =
(D)
x
+c
1 log x
– Sol.[C] put log x = t, x = et, dx = et dt
16
t 1 

 1  t 2  2t  4 


 t
 dt



 e  log t 

dt
t2 
 1 1 1


1 5 1
  2  t  t  dt
t
= – = –  e log t     dt [et (f(t) + f '
16 16  t t t2 
 t 2  (t))]
2t   5 log | t | + 
 1
 2  = et  log t   + c =
1 1 5  t
=– cos2 2x – cos 2x – log |cos 2x| +
32 8 16  1 
x log(log x )   +c
  log x

x
 x.( x
x
Q.154 ) (2 log x + 1) dx
x x
Q.157  (sin 2x  cos 2x )dx =
(A) ( x ) + c (B) log (x)x + c
(C) xx + c (D) None 1
x
Sol.[A] put ( x x ) = t  x log xx = log t sin(2 x  a )  b , then-
2
 x2 log x = log t

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 x x
5 5 2 3
(A) a =
4
, b  R (B) a = –
4
,bR Sol.[B]
      dx
5 5
=

 ( 2 / 5) x (3 / 5) x
(C) a = ,bR (D) none of these [B]  +C
4 log(2 / 5) log(3 / 5)

d (cos ) Q. 163  tan 2x tan 3x tan 5x dx is equal to -

Q.158  1  cos 2 
=
1 1 1
(A) log|sec 2x| – log|sec 3x| – log|sec
2 3 5
(A) cos–1 + c (B)  + c
(C) sin–1 + c (D) sin–1(cos) + c [D] 5x|+C

2x 1 1 1
Q.159 If  1 4x
dx = K sin–1(2x) + C then K is (B)
2
log|sec 2x|+
3
log|sec 3x|+
5
log|sec 5x|

+C
equal to-
1 1 1
1 (C) log|sec5x|– log|sec 2x|– log|sec
(A) log 2 (B) log 2 5 2 3
2
1 1 3x|+C
(C) (D)
2 log 2 (D) None of the above
[D] tan 2 x  tan 3x
Sik.[C] tan 5x = tan (3x + 2x) =
Q.160 Evaluate  {{[ x ]}} dx , where {  } and [  ] 1 – tan 2 x. tan 3x
 tan 5x – tan2x. tan 3x. tan 5x = tan 2x + tan 3x
denotes fractional part of x and greatest integer   tan2x. tan3x. tan5x dx =  tan5x – tan2x – tan3x.
function of x dx
Now solve.........
(A) 0 (B) 1 (C) 2 (D) – 1 1
x x
Q.164  (x  1) x2 –1
dx is equal to -
2 3
Sol.[B]
      dx
5
  5
=
(A)
1 x –1
+ C (B)
x 1
+C
( 2 / 5) x
(3 / 5) x 2 x 1 x –1
 +C
log(2 / 5) log(3 / 5) 1 x 1 x –1
(C) +c (D) +C
Q.161 The value of  [{x}] dx , where [  ] and { 2 x –1 x 1

 } denotes greatest integer and fractional part of 1 1

Sol.[D] Let x + 1 = t then dx = – t 2 dt
x is equal to 1
(A) 0 (B) 1 (C) 2 (D) – 1
1 1 
2  1
– 2 
I =  – 1 – 1 ×  t  dt = –
t  t 
2 x  3x
Q.162
 5x
dx is equal to -

dt
1 – 2t
( 2 / 5) x (3 / 5) x (1 – 2 t )1 / 2 x –1
(A)  +C = –  (1 – 2 t ) –1 / 2 dt = – +C =
log 3 (3 / 5) log 3 (2 / 5) (–2)(1 / 2) x 1
(2 / 5) x (3 / 5) x +C
(B)  +C
log 3 ( 2 / 5) log 3 (3 / 5)
sin 4 x
(C)
(3 / 5) x

(2 / 5) x
+C
Q.165
 cos 8
x
dx is equal to -
log 3 (2 / 5) log 3 (3 / 5)
(D) None of the above (1  tan 5 x ) tan 5 x
(A)  +C
5 7
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 tan 5 x tan 7 x 1 1/ m
 C =  (2x d (2x3m + 3x2m +
3m
(B)  3x 2 m  6 x m )
5 7 6m
6xm)dx
tan 7 x tan 5 x
(C)  +C
5 7 1
– (2x3m + 3x2m + 6xm)1/m+1 + C
6( m  1)
(D) None of these
d (cos )
sin 4 x Q.168  1 – cos 2 
d
cos 4 x .dx =  tan4x. sec4x dx
Sol.[B]  cos 8 x (A) cos–1 + c (B)  + c
cos 4 x
(C) sin–1 + c (D) sin–1(cos ) + c
=  tan4 x (1 + tan2x)sec2x dx =  tan4x.
t5 t7 d (cos )
(1 + tan2x) d(tanx) = t4(1 + t2)=
5
+
7
+ Sol.[D]  1 – cos 2 
d
C
sin  d
1  sin x cos x 
=–  1 – cos 2 
Q.166  ex 
 cos 2 x
 dx is equal to -
 put t = cos 
dt = – sin  d
(A) ex cos x + c (B) ex sec x . tan x + c
dt
(C) ex tan x + c (D) ex cos2x –1 + c =  1– t2
x
1 sin x. cos x 
Sol.[C] 
e  2
 (cos x )

cos 2 x 
 = sin–1t + c
= sin–1 (cos ) + c
=  e { sec2x + tanx} = ex tan x + c
x

Q.167 If m any natural number, then the value of the 1 1

Q.169 lim + + ......
n  2
2n – 1 4n – 2 2
integral  ( x
3m 2m m
x x ) (2x 2m m
+ 3x + 6) 1/m
1 1
2
+ .........+ is equal to :
dx is – 6n – 3 n
1  
(A) (B)
(A) {2x3m + 3x2m + 6xm}(1/m)+1 + C 4 2
6( m  1)
 
(C) (D)
1 6 3
(B) {2x3m + 3x2m + 6xm} (1/m)+1 + C
6m 1 1
Sol.[B] lim + +
n  2
1 2n – n 2 .2 n – 2 2
(C) {2x3m + 3x2m + 6xm}1/m + C
6m 1
(D) None of the above 3.2n – 32
Sol.[A]  ( x3m  x 2 m  x m) 1/ m 1
3m 2m m

 ( 2 x  3x  6 x )  + ..........
  .dx n.2n – n 2
 x 
  n
1

dx
( x 3m –1  x 2 m –1  x m –1 ) (2x3m +3x 2m m 1/m
+6x ) lim
n  
r 1 2rn – r 2
1 n
1

1/ m
 (2 x (6x3m–1 + 6x2m–1 +
3m
 3x 2 m  6 x m )
6 lim 2
n  r 1 r r
6xm–1)dx n 2 – 
n n

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 1
 
x1 
dx
=  (put r/n = x)
1 4
tan–1 
0 2x – x 2
I= 4 +C
1 2 7  7 
dx  
= 
0 1 – ( x – 1) 2  4 
2  4x  1 
= sin 
1  I= tan–1  +C
7 
–1
( x – 1) = 7
0
2 

x
x
Q.172 (1  log x ) dx =
cot x
Q.170
 sin x cos x
dx = P cot x + Q, then P x
(A) x log x + C (B) xx + log x + C
(C) xx + C (D) log (xx) + C [C]
equals to :
(A) 1 (B) 2 (C) –1 (D)
dx
–2 Q.173 If  ( x 2  1) (x 2  4)
cot x dx x
Sol.[D]
 sin x cos x = K tan–1 x + L tan–1
2
+ C, then

divide by sin2x 1 2
(A) K = (B) L =
cot x 3 3
 cot x
.cosec2x dx
(C) K = –
1
(D) L =
1
3 6
t
 t
. (– dt) (t = cot x) [A]

= –2 t +c sin 2 x
=–2 cot x + c
Q.174 If  sin 5x sin 3x dx = A log |sin3x| +

B log |sin 5x| + C. Then

1 1 1 1
dx (A) A = ,B= (B) A = – ,B=
Q.171  2x 2
 x 1
equals : 3
1
5
1
3 5
1
(C) A = ,B=– (D) A = ,
2  4x  1  3 5 3
(A) tan 1 C
7  7  B=–
1
[D]
5
1  4x  1 
(B) tan 1  C
2 7  7  e x (1  sin x )
Q. 175
 1  cos x
dx =
1 1  4 x  1 
(C) tan  C x
2  7  (A) log |tan x| + C (B) ex tan   + C
2
(D) None (C) sin ex cot x + C x
(D) e cot x + C [B]
dx
Sol.[A] I =
1 x 2 x 1
 
Q. 176 The value of  cos (log x ) dx is:
2 1
2 2 (A) [sin (log x) + cos (log x)] + C
dx 2
I=
1  1
2
 7
2
x
2  x     
 (B) [sin (log x) + cos (log x)] + C
 4  4  2
x
(C) [sin (log x) – cos (log x)] + C
2

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 1 sin 8 x  cos 8 x
(D)
2
[sin (log x) – cos (log x)] + C [B] Q.182  1  2 sin 2 x cos 2 x dx is equal to-
1 1
3
(A) sin 2x + c (B) – sin 2x + c
x 2 2
Q. 177
 x  1 dx = (C) –
1
sin x + c (D) – sin2 x + c [B]
2
x3 x4 10x 9  10 x log e 10
(A)
3
+
4
+C Q.183
 x10  10 x
dx equals-

x3 x2 (A) 10x – x10 + C (B) 10x + x10+ C

(B) – + x – log (x + 1) + C
3 2 (C) (10x – x10)–1+ C (D) log (10x + x10) + C [D]
x3 x2 Q.184
(C) – + x + log (x + 1) + C
3 2
1
(D)
x4
+C [B]  xl (x )l 2
( x )l ( x ).....l n 1 ( x )l n ( x )
3
dx
2( x  1) 2
, is equal to :
(1  x )
Q. 178  x (1  xe x ) 2 dx is where ln (x) = loge

 

 log e (log e ....(log e x ).....) 

x 1   n  times  
xe
(A) log + +c
1  xe x 1  xe x (A) l n (x) + c (B) l n + 1 (x) + c
x
xe (C) l n – 1 (x) + c (D) None of these [B]
(B) (1+ xex) + log x +c
1  xe
1
(C) + log |xex(1+ xex)| + c 1
1  xe x
(D) None of these [A]
Q.185  e5x 4
(e 2 x  e  2 x ) 3
dx is equal to :

1
x  x 2 / 3  x1 / 6 (A) – (1 + e–4x)1/4 + c (B)
Q. 179 I =  x (1  x1 / 3 )
dx is equal to- 4

3 2/3 (1  e 4 x )1/ 4
(A) x + 6 tan–1 (x1/6) + c  c
2 1/ 4
3 2/3 (C) – (1 + e–4x)1/ 4 + c (D) None of these [C]
(B) x – 6 tan–1 (x1/6) + c 49 1 50
2 x tan (x )
3 2/3
Q.186  (1  x 100
)
dx  k(tan–1(x50))2 + C,
(C) x + tan–1 (x1/5) + c
2 then k is equal to
(D) None of these [A]
:
2
1 (1  x  x )
Q. 180 The value of  e tan x dx is 1 1
1 x2 (A) (B) 
50 50
(A) xe tan 1 x + C (B) tan–1 x + C
1 1
(C) (D) 
(C) e tan 1 x + 2x + C (D) None of these [A] 100 100
[C]
x  sin x
Q. 181 
1  cos x
dx is equal to-
Q.187
x  sin x
 1  cos x dx is equal to
x x
(A) x tan +C (B) cot +C x x
2 2 (A) – x cot +c (B) cot +c
(C) log (1+ cos x) + C (D) log (x + sin x) + C [A] 2 2

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 x 1 –2x
(C) – cot +c (D) None of these [A] (A) e [sin 3x + cos 3x ] + c
2 13
x2 1 –2x
Q.188  (x sin x  cos x ) 2
dx is equal to (B) –
13
e [sin3x + cos 3x] + c

x sec x 1 –2x
(A) + tan x + c (C) e [2sin 3x + 3 cos 3x] + c
x sin x  cos x 13
 x sec x 1 –2x
(B) + tan x + c (D) – e [2sin 3x + 3 cos 3x] + c
x sin x  cos x 13
e
ax
Sol.[D] Use formula . sin( bx  c) dx
 x sec x
(C) + sec x tan x + c
x sin x  cos x e ax
= [a sin(bx + c) – b cos(bx + c)] + c
 x sec x a 2  b2
(D) + 1 – tan2 x + c
x sin x  cos x
[B] x
3
Q.193 logx dx =
2
(1  x )
e x 4 log x 1
x
Q.189 2
dx is equal to (A) +c (B) [4x4logx– x4] + c
(1  x ) 16
4
1 ex 1 4 1
(A) c (B) c (C) [x logx – 4x2] +c (D)
x2 1 x2 1 8 16
ex 1 1 ex [4x4 logx + x4] + c
(C) c (D) c [B]
x2 1 x2 1 Sol.[B]  x logxdx=logx  x dx
3 3

dx
Q.190  x(x n
 1)
is equal to :  d
  dx (log x ). x dx dx
3 

1  xn  x4 1 x4 x4 1
log n +C
(A)
n 
 x 1 
(B) =
4
logx –
 .
x 4
dx =
4
log x –
4
 x dx
3
1  xn 1
log n
 +C

n  x  x4 1 4 1
= log x – x + c  [4x4 log x –
 xn 1 4 16 16
(C) log n
 +C (D) None of these
 [A] x4] + c
 x 
d
Q.194 If f(x) = x cosx + sinx and f(0) = 2, then f(x)
dx
sin 2 x
Q.191  sin x  cos 4 x
4 dx = =
(A) x sin x (B) x cosx + sinx + 2
(A) cot–1(tan2x) + c (B) tan–1(tan2x) +c (C) x sinx + 2 (D) x cosx + 2
(C) cot–1(cot2x) + c (D) tan–1 (cot2x) + c Sol.[C] Integrating both sides :
2 sin x cos x 2 tan x sec 2 x d
Sol.[B]  sin 4 x  cos 4 x
dx =
 1  tan 4 x
 dx f(x) =  x cos xdx   sin xdx
dx f(x) = x  cos x dx –  1. cos xdx  dx +
Put tan2x = t 2tanx. sec2x dx = dt
dt  sin x dx
  1 t2
= tan–1(t) + c = tan–1(tan2x) + c f(x) = x sinx –  sin x dx +  sin x dx
f(x) = x sin x + c f(0) + 2
2 = 0 + c  c = 2  f(x) = x sin x + 2
e
–2 x
Q.192 sin 3x dx =

e
x
Q.195 (tanx – log(cos x)) dx =
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 (A) ex log (sec x) + c (B) ex log (cosec x) + c (C) 2  e x – 1  tan –1 e x – 1  c


(C) ex log (cos x) + c (D) ex log (sin x) + c
–1
(D) e x – 1 + tan ex – 1 + c
e
x
Sol.[A] use (f(x) + f '(x)) dx = ex . f(x) + c
Sol.[A] Put ex–1 = t2 ex dx = 2t dt
 sin 2 t dt 2t

–1
Q.196 (3x – 4 x 3 ) dx 
dx = 2  t. 2 dt
t 1 t 1
–1
(A) x sin x + 1– x 2 +c
2t 2 2t 2  2 – 2
(B) x sin–1x – 1 – x2 + c
=
 t2 1
dt =
 t2 1
dt

2( t 2  1) 1
(C)2[x sin–1x + 1 – x 2 ] + c =
 2
t 1
dt – 2 2
t 1
dt
–1
(D) 3[x sin x + 1 – x 2 ] + c = 2  dt – 2 tan –1 ( t )  c = 2t – 2 tan–1(t) + c
Sol.[D] Put x = sin = 2 e x – 1 – 2 tan–1( e x – 1 ) + c
dx = cos d
 sin (3sin – 4 sin3) cos d
–1

d2
=  sin (sin 3) . cos
–1
d= Q.200
 dx 2
(tan–1x) dx is equal to -
3 . cos .d
1
= 3 [  cos  d – {d (). cos d}d]
d (A) +c
1 x2
= 3 [ sin  –  sin  d] = 3[ sin  + (B) tan–1x + c
cos] + c 1
(C) x tan–1x – log (1+x2) + c
= 3[x sin–1x + 1– x 2 ]+c 2
(D) None of these

x
x d  d 
Q.197 . (1 + logx) dx =
 dx  dx (tan
–1
Sol.[A] x ) dx
(A) xx (B) x2x 
1 d 1
(C) xxlogx (D) (1 + log x)2 = tan–1x + c = +c
2 dx 1 x2
Sol.[A] Put xx = t
sin x
xx(1+ logx) dx = dt   dt = t + c = xx + c Q.201 The value of  x
dx is

(A) 2cos x +C (B) – 2 cos x +C

cos 4 x  1
Q.198 If  cot x – tan x
dx = k cos 4x + c, then (C) cos x +C (D) – cos x +C
Sol. Let x =t
(A) k = – 1/2 (B) k = – 1/8
(C) k = – 1/4 (D) None of these 1
dx = 2dt
2
x
2 cos 2 x
Sol.[B]
 cos
2
x – sin 2 x
. cos x. sin x dx
 2  sin t dt = 2 cos x +C
1
=  cos 2 x sin 2x dx = – cos4x + c  k = – 1
1/8
8 Q.1 The value of  1  5x dx is

(A) log (1–5x) + C (B) – log (1 – 5x) + C

1
Q.199  ex – 1 dx (C) – 5 log (1– 5x) + C (D) 
5
log (1 – 5x) + C
 e x – 1  c
(A) 2 
e x – 1 – tan –1
 n (1  5x )
 
Sol. C
–1
( 5)
(B) e x – 1 – tan ex – 1 + c
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 sin x
Q.90 If  sin( x  ) dx  Ax + B log sin(x – )
3
x2
Q.1 dx is equal to
+ c, then value of (A, B) is –
3x  2
(A) 3x+2 + c (B) +c
log 3 (A) (– sin , cos ) (B) (cos , sin )
x 2
3 (C) (sin ,cos ) (D) (– cos , sin ) [B]
(C) +c (D) none
log 2
3x  2
Sol. +c (tan x )
log 3 Q. 92  sin x cos x dx =

(A) 2 (tan x ) + c (B) 2 (sec x ) + c

Q.2 Evaluate  1  cos 2 x dx
(C) 2 / (tan x ) + c (D) None of these [A]
(A) 2 sin x + c (B) 2 cos x + c
(C) 2 tan x + c (D) none
2 x  1  sin 2 x
Sol.  2 cos 2
x dx = 2  cos x dx Q.98 e   dx =
 1  cos 2 x 
= 2 sin x + c 1 2x 1 2x
(A) e cot x + c (B) e tan x + c
2 2
1 2x
e x (1  sin x ) (C) – e cot x + c (D) None of these [B]
Q.86
 1  cos x
dx is equal to – 2
1
(A) log tan x (B) ex tan x Q.99 The value of constant of integration which
2 vanish the integral of sin 3x cos 5x at x = 0 is
(C) sin log x (D) ex cot x. [B] given by-
(A) 3/16 (B) –3/16
Q.87 If x sin x dx = – x cos x + , then  = (C) 1/8 (D) 0 [B]
(A) sin x + c (B) cos x + c
(C) x cos x + c (D) cos x – sin x + c. [A]
 tan
3
Q.100 2 x sec 2 x dx =
(A) sec32x + 3 sec 2x + c
e
3 log x 4 1
Q.88 (x  1) dx 
1
(B) [sec32x – 3 sec 2x] + c
(A) log (x4 + 1) + c (B) 3 log(x4 + 1) + c 6
(C) sec3 2x – 3 sec 2x + c
1
(C) – log(x4 + 1) + c (D) log(x 4  1)  c (D) None of these [B]
4
[D]
Q.101 Integral of f(x) = (1  x 2 ) with respect to
x
Q.89  a3  x3
dx  g ( x )  c , then g(x) = x2 is-
2 (1  x 2 ) 3 / 2 2
(A) +k (B) (1 +
2 2  x3  3 x 3
(A) cos –1 x (B) sin  3  x2)3/2 + k
3 3 a 
  2
2 (C) x(1 + x2)3/2 + k (D) None of these [B]
(C) sin –1 ( x 3 / a 3 ) (D) 3
3
2 Q.102 If r means (log log log......), the log being
cos –1 ( x / a ) [C] repeated r times, then
3
 {x( x ) ( x ) ( x ).... ( x )} dx is
2 3 r 1

equal to
r 1 ( x )
(A) r+1 (x) + c (B) +c
r 1
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 (C) r (x) + c (D) None of these [A] x 49 tan 1 ( x 50 )
Q.112  (1  x100 )
dx  k(tan–1(x50))2 + C,
Q.103  sin 2x log cos x dx =
then k is equal to
1 
(A) cos x   log | cos x |  + k
2 :
2 
1 1
(B) cos2 x log |cos x| + k (A) (B) 
50 50
1 
(C) cos2 x   log | cos x |  + k 1 1
2  (C) (D) 
(D) None of these [C] 100 100
[C]
3
cos x  cos x
Q.104
 1  cos 3 x
dx =
Q.133
x  sin x
 1  cos x dx is equal to
(–/2 < x < 0)
2 x x
(A) – sin–1(cos3/2x) + c (A) – x cot +c (B) cot +c
3 2 2
3 x
(B) sin–1 (cos3/2 x) + c (C) – cot +c (D) None of these [A]
2 2
2
(C) cos–1 (cos3/2x) + c
3 x2
(D)
2
sin–1(cos3/2x) + c
Q.134  (x sin x  cos x ) 2
dx is equal to
3
[D] x sec x
(A) + tan x + c
x sin x  cos x
Q.110  x sec x
(B) + tan x + c
x sin x  cos x
1
 xl (x )l 2
( x )l ( x ).....l n 1 ( x )l n ( x )
3
dx
(C)
 x sec x
x sin x  cos x
+ sec x tan x + c

, is equal to :  x sec x
(D) + 1 – tan2 x + c
where l n
(x) = loge x sin x  cos x
[B]
 

 log e (log e ....(log e x ).....) 

  n  times   x5
(A) l n (x) + c (B) l n + 1 (x) + c
Q.85  1 x2
dx =

(C) l n – 1 (x) + c (D) None of these [B] 1 4 2

(A) 1  x 2 (3x + 4x + 8)
15
1
Q.111  e 5x 4
(e 2x
 e 2x )3
dx is equal to :
(B)
1 4 2
1  x 2 (3x – 4x + 8)
15
1
(A) – (1 + e–4x)1/4 + c (B) (C) 4 2
1  x 2 (3x + 4x + 8)
4
(D) none of these [B]
(1  e 4 x )1/ 4
 c
1/ 4
 (1  x – x
–1

(C) – (1 + e–4x)1/ 4 + c (D) None of these [C] Q.86

–1
) ex x dx =

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 (A) (x + 1) e x  x
–1
+C (B) (x – 1) (D) None of these [B]

–1
ex x +C Q.91 If the equation of a curve passing through the
–1 –1
(C) – xe x  x +C (D) xe x  x + C [D]
x
3
2
point (0, 1) is given by y = . e x dx and

the equation of the curve is written as

Q.87 If In =  (log x ) n dx, then In + nIn – 1 = x = f(y) then f(y) =

(A) (x log x)n (B) x ( log x)n (A) log e (3y – 2) (B)
n n –1
(C) n (log x) (D) (log x) [B]
3 log e (3y – 2)

Q.88 If  g ( x ) dx  g ( x ), (C) 3 log e ( 2 – 3y ) (D) None of these [B]

then  g ( x ) (f ( x )  f ' ( x )) dx =
dx
(A) g(x) f2(x) Q.92 x 3
( x 3  1)1/ 3
=

(B) g(x) f(x) – g(x) f '(x)

1
(C) g(x) f''(x) (A) – (1 + x3)2/3 + C (B) – (1 + x–3)2/3 + C
2
(D) g(x) f(x) [D]
(C) – (1 + x3)–2/3 + C (D) None of these [B]

Q.89 If f(x) is a polynomial satisfying

1 1
f(x) . f   = f(x) + f   , and f(3) = 82, Q.93  (log (log x )  (log x )
–1
) dx =
x x
then (A) x log (log x) + C
x
f (x )
x 2
1
dx =
(B) x log (log x) +
log x
+C

x
(A) x3 – x + 2 tan–1 x + C (C) x log (log x) – +C
log x
1 3
(B) x – x + tan–1 x + C (D) None of these [A]
3

x3
(C) – x + 2 tan–1 x + C 
3 Q.94 The primitive of x | cos x | when < x <  is
2
1 3 -
(D) x + x + 2 tan–1 x + C. [C]
3
(A) cos x + x sin x (B) – cos x – x sin x

f (x) (C) x sin x – cos x (D) None of these [B]

Q.90 If f ' (x) = f(x) and f(0) = 2, then  3  4f ( x )
e
x log a
dx = Q.95 . ex dx =
(A) log (3 + 8 ex) + C
(ae) x
1 (A) (ae)x (B)
(B) log (3 + 8 ex) + C log (ae)
4
1 ex
(C) log (3 + 8 ex) + C (C) (D) None of these [B]
1  log a
2

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 e x (1  sin x ) 5 3
Q.96  1  cos x
dx = (B) x5 –
3
x + 5x +c

(C) 3x4 – 5x2 + 15x + c

1 1
(A) ex cot x+C (B) ex tan x+C (D) 5tan–1 (x2 + 1) + log(x2 + 1) + c
2 2
5( x 6  1)
(C) ex tan x + C (D) ex cot x + C [B] Sol.[B]
 x2 1
dx =

dx
Q.97  sin x  cos x =
5( x 2  1)( x 4 – x 2  1)
(A)
1  x 
log tan    + C
 ( x 2  1)
dx

2 2 8
5 3
 5( x x + 5x + c
4
 x  = – x 2  1)dx = x5 –
(B) log tan    + C 3
2 8
ax –2  bx –1  c
(C)
1
2
 x 
log tan    + C
2 8
Q. 69
 x –3
dx =

(A) 2ax2 + 3bx3 + 4cx4 + k

1   (B) 6ax2 + 4bx3 + 3cx4 + k
(D) log tan  x   + C [A]
2  4 (C) a + b + cx2 + k
1 2 1 3 1 4
(D) ax + bx + cx + k
x –1 2 3 4
Q.98  ( x – 3)(x – 2) dx =
ax –2  bx –1  c
(A) log (x – 3) – log (x – 2) + C
Sol.[D]
 x –3
dx =

(B) log (x – 3)2 – log (x – 2) + C

 (ax  bx
2
 cx 3 )dx

(C) log (x – 3) + log (x – 2) + C

1 4 1 3 1 2
(D) log (x – 3)2 + log (x – 2) + C [B] = cx  bx  ax  k
4 3 2

dx
Q. 70  sin x  3 cos x
=

Q. 67 a
x
da   x 
(A) log tan   +c
2 2
ax
(A) c (B) ax logea + c 1  x 
log e a (B) log tan    + c
2 2 6
a x 1  x 
(C) c (D) xax–1 + c (C) log cot    + c
x 1 2 6
a x 1  x 
Sol.[C]
 a x da 
x 1
c (D)
1
2
log cot    + c
2 6

5( x 6  1)
Q. 68
 x2 1
dx =

(A) 5(x7 + x)tan–1x + c

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 dx 1 1
Sol.[B]  sin x  3 cos x
=
2
=
6
log (sec 6x + tan6x) + c

sin 3x
Q. 79  sin x
dx =
dx
 sin x 3
(A) x + sin2x + c
(C) 3x + sin2x + c
(B) 3x + sin 2x + c
(D) None of these
 cos x
2 2 sin 3x 3 sin x – 4 sin 3 x
1 dx
Sol.[A]  sin x
dx =
 sin x
dx

= 2  sin x    =
1

 
cos ec x    3dx – 4  sin
2
xdx = 3x –2
  2  3
 3  (1 – cos 2 x ) dx +c
= 3x – 2x + sin2x + c = x + sin2x + c
1  x 
= log tan    c f ( x )dx
2 2 6 Q. 80 If  log sin x = log log sin x, then f(x) =

x2  x – 6
Q. 77  ( x – 2)( x – 1)
dx =
(A) sin x (B) cos x
(A) x + 2log(x – 1) + c (C) log sin x (D) cot x
f ( x )dx
(B) 2x + 2log(x – 1) + c
(C) x + 4log(1 – x) + c
Sol.[D]  log sin x = log log sinx

(D) x + 4log(x – 1) + c
Differentiating both sides, we get
x2  x – 6 ( x  3)( x – 2)
Sol.[D]  ( x – 2)( x – 1)
dx =  ( x – 2)( x – 1) f (x) cot x
=  f(x) = cotx.
log sin x log sin x
dx
1
= 
x 3
dx
Q.87 x 2
(2 x  1)3 dx =
x –1
1
(A) 4x2 + 12x + 6logx – +c
x –1 4 x
=  x – 1 dx   x – 1 dx = x + 4log (x – 1) (B) 4x2 + 12x – 6logx –
2
+c
x
+c 2
(C) 2x2 + 8x + 3logx – +c
x
2 2
dx (D) 8x + 6x + 6logx   c
Q. 78  4 cos 3
2 x – 3 cos 2 x
= x
1
(A)
1
log[sec6x + tan6x] + c
Sol.[A] x 2 (2x + 3
1) dx =
3
1
(B) log[sec6x + tan6x] + c (8x 3  1  12 x 2  6 x )
6
(C) log[sec 6x + tan 6x] + c
 x2
dx

(D) None of these  6 1 

dx dx
=   8x  12  x  x 2  dx
Sol.[B]  3
4 cos 2 x – 3 cos 2 x
=  cos 6 x
= 
1
 sec 6 x dx = 4x2 + 12x + 6 logx –
x
+c

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 (ae) x
e 5 log x – e 4 log x  (ae)
x
= dx = + C.
Q.88
 e 3 log x – e 2 log x
dx = log(ae)

(A) e.e–3x + c (B) e3logx + c

x3
Q.81 If  2 1  sin x dx = – 4 cos (ax + b) + C,
(C) c (D) None of then (a, b) =
3
1  
these (A) , (B) 1,
2 4 2
e5 log x – e 4 log x x5 – x4 (C) 1, 1 (D) none of these
Sol.[C]
 e3 log x – e 2 log x
dx =
 x3 – x2
dx
Sol. [A] 2  1  sin x dx =
 x x
= 
x 4 ( x – 1)
dx =

2
x dx 
x3
c 
2  cos  sin  dx
 2 2
x 2 ( x – 1) 3
x 
x4 – x2 1
= 2  sin  2  4  dx = 2
Q.89
 x2 – x 1
dx =
x 
1 3 1 2
 – cos 2  4  +C
(A) x  x xc (B) 1
3 2
2
1 3 1 2
x – x xc   x  
3 2 = – 4 cos     C
  2 4 
1 3 1 2
(C) x  x – x  c (D) None of
3 2
1– x
these Q.82
 1 x
dx =
x4  x2 1
Sol.[A]
 x2 – x 1
dx
(A) sin–1 x –
1
1– x2 + c
2
 (x
2
=  x  1)dx 1
(B) sin–1 x + 1– x2 + c
2
x3 x2 (C) sin–1 x –
=  xc 1– x2 + c
3 2
(D) sin–1 x + 1– x2 + c
1– x
e 
x log a
Q.90 .ex dx is equal to -
Sol. [B] 1  x dx , multiply and divide by
1 x
 ae  x 1– x
(A) (ae)x + c (B)
log(ae)
c
= 1– x2
 dx =
dx x
(C)
ex
c (D) None of these  1– x 2
–
1– x2

dx
1  log a
(–2 x )
Sol.[B] e
x log a
e x dx = e
log a x
.e x dx = –1
1
= sin x + 2 1 – x 2 dx

1 2
= sin–1x + 2  2 1 – x + c
a
x x
e dx

a
3 x 3
Q.83 dx =

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 a 3x  3 a 3x  3 dx
(A)
log a
c (B)
3 log a
c Sol. [D]  sin x – cos x  2
3x + 3 3x + 3
(C) a log a + c (D) 3a log a + c dx
  = 
Sol. [B] a 3x
 a 3
dx = a 3
a3xdx =  1 
1
a 3x 2  sin x – cos x   2
a3   2 2 
3 log a
a 3x  3 1 dx
2) 
c
= 3 log a = (–  
cos x   – 1
x
 4
5
55 5x x
Q. 84 5  5 dx 

1 dx
2) 
x
55 5x
(A) C (B) 55 3
(log 5) + C = (   =
(log 5) 3 1 – cos x  
 4
5x
55
(C) C (D) none of these
1 dx
2
(log 5) 3
x 
Sol. [C] Do yourself 2 sin 2   
2 8
log( x  1) – log x
Q. 85  x ( x  1)
dx is equal to - 1 2x 
2
= cos ec   dx =
2
2 2 8
1   x  1 
(A) – log   C
2   x 
 x 
(B) C – [{log (x + 1)}2 – (log x)2] cot  dx
  x  1  –  2 8 C
(C) – log log
  x 
  C
 1
2 2  
2

 x 1
(D) – log  +C
 x  1  x 
= – cot    + C
Sol. [A] Put log (x + 1) – log x = t 2 2 8
1 1 dt x – ( x  1) dt
–  
 x  1 x dx  x ( x  1) dx Q. 87  [1 + tan x  tan (x + )] dx is equal to -
– dx dx
sin x
 x ( x  1) = dt  x ( x  1) = – dt (A) cot . log C
sin( x   )
so question becomes
t2 sin x
–  t dt = – 2  C (B) tan  . log
sin( x   )
C

dx
Q. 86 
sin x – cos x  2

(C) cot  . log
cos x
C
cos( x   )
1  x  (D) none of these
(A) – tan     C
2 2 8
 sin x sin( x  ) 
(B)
1 x 
tan     C
Sol. [C]  1  cos x  cos(x  )  dx
2 2 8
=
1 x 
(C) cot     C
2 8  cos x cos( x   )  sin x sin( x   ) 
2
  cos x. cos( x   )
 dx

1  x 
(D) – cot     C
2 8 cos( x    x )
2
=  cos x. cos(x  ) dx
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 dx x
= cos   cos x. cos(x  ) =
2 sin   dx
2 = –2dt
Multiply denominator and numerator by sin  So Integral becomes
dt
=
cos 

sin 
dx =
2 
t 2  1 = –2 log
t 2 1  t
+C
sin  cos x. cos( x  )
sin( x    x )
= cot   cos x. cos( x  ) dx Q.89 If  (1  x )
1
x
dx = f (x) + A, where A is
= any arbitrary constant, then the function f (x) is
sin( x  ) cos x  cos(x   ) sin x -
cot   cos x. cos(x   )
dx
(A) 2 tan–1x (B) 2 tan–1 x
–1
(C) 2 cot (D) loge (1 + x)
= 
cot  {tan( x  )  tan x} dx x

cot  log | sec(x  ) |  log | sec x |  C

Sol.[B] Put x =t
=
1 dt dx
sec( x   ) 2 x = dx  x = 2dt
= cot  log C
sec x 1
cos x
2  1 t 2
dt
= 2 tan–1t + C
= cot  log C
cos( x   ) cos x
Q.90 If  sin(x – ) = A log sin (x – ) – Bx +

Q. 88  sec x – 1 dx is equal to - C then (A, B) =

(A) (cos , sin ) (B) (sin , cos )
 x 2 x 1
(A) 2 log  cos  cos – C (C) (– cos , sin ) (D) ( – sin , cos )
 2 2 2 
cos( x     )
 x
(B) log  cos  cos
2 x 1
– C
Sol.[A]  sin( x   )
dx

 2 2 2  =
cos(x  ) cos   sin( x  ) sin 
 x
(C) – 2 log  cos  cos
2 x 1
– C
 sin( x  )
dx

 2 2 2 
= cos   cot( x  ) dx  sin   dx
(D) none of these
1 1  cos x
Q.91  sin4x dx, = ax + b sin 2x + c sin 4x + d
Sol.[C]  cos x
 1 dx
=  cos x
dx then 8a + 4b + 32 c =
(A) 0 (B) 1
x
2 sin 2   (C) 2 (D) none of these
 2  dx
 x
2 cos 2    1 Sol.[D]  sin 4 x dx
= 
 1  cos 2 x 
 
2
dx
= =  2 
2 1
 (1  cos
2
x = 4 2 x  2 cos 2 x ) dx
sin  
2
2  2 x 
dx x
= 4 – 2. 2
1 sin 2 x

4 2
1

1  cos 4 x
dx
2 cos    1
2 x sin 2 x x sin 4 x
–   C
= 4 4 8 32
x
2 cos  3x sin 2 x sin 4 x
Put 2 =t –  C
= 8 4 32
x 1 dt
2 sin   .
=– 2 2 = dx
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     (f ( x )) 2 x  x sin 2 x 
Q.92  sec x –  sec x –  dx =
 3  6  2 =
 –
2 – 2 4 
C
sin 2 x
cos( x –  / 3)
(A) log C (f(x))2 = 2 + C, f(x) =
cos( x –  / 6)
| sin 2 x |
cos( x –  / 3) C
(B) 2 log C 2
cos( x –  / 6) 2
cos( x –  / 6) period of f(x) is 2 = 
(C) 2 log C
cos( x –  / 3) dx
cos( x –  / 6)
Q.94  x 2  2x  1
= A log |x + 1| + C for x < –
(D) log C
cos( x –  / 3) 1 then A is -
dx (A) 0 (B) 1 (C) – 1 (D) None
Sol.[B]      dx dx
cos x –  cos x – 
 3  6 Sol 
( x  1) 2 = | x  1 | = log|x + 1| + C
1
=  
sin  / 6 3 4
1  x  x  x  ..... dx equals to-
     
Q.105  
 3 4 

sin  x –  –  x – 
  6  3  1 x
   
cos x –  cos x – 

dx (A) ex + c (B)
2
e +c

 3  6 (C) 2ex + c (D) None of these

= 2  x3 x4 
        Sol.[D]
  1  x  3

4
 ..... dx

sin x –  cos x –  – cos x –  sin  x –  
 6  3  6  3
     
cos x –  cos x – 

dx
= 
 x x2
e 
 2


6
3
dx = ex – x + c
 
 3  6
     
=2  tan x –  – tan x – dx
  6  3  Q.106 
d ( x 2  1)
=
= 2 x2  2

      
 (A)
 – log cos x –   log cos x –    C x2  2 + c (B) 2 x 2  2 + c

  6   3 
 (C) (x2 + 2)–3/2 + c (D) None of these
 
cos x –  d ( x 2  1) 2x
 3 Sol.[B]  x 2
2
dx =  x2  2
dx = 2
 
= 2log cos x –  +C
 6 x2  2 + c
1
– sin 2 x
Q.93 If the derivative of f(x) w.r.t x is 2 sec 2 x tan x
f (x)
Q.107
 sec 2
x  tan 2 x
dx =

then period of f(x) is- (A) log(sec2 x + tan2 x) + c

(A) 2 (B)  (C) /2 (D) None
1 (sec 2 x  tan 2 x ) 2
(B) +c
– sin 2 x 1 2
Sol.[B] f (x) = 2  f(x) f  (x) = –
2
f (x) 1
2
(C) log (sec2 x + tan2x) + c
sin x 2
 (D) None of these
1 
 f (x )f (x )dx    2 – sin
2
x dx

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 sec 2 x tan x x
x
Q.110 (1 + log x) dx is equal to-
Sol.[D]
 sec 2
x  tan x 2
dx t =
(A) xx log x + k (B) e x
x
+k
sec2 x + tan2x
dt / 4 dt (C) xx + k (D) None of these
=  t dx
=2secx.secx tan x
Sol.[C] x
x
(1  log x )dx = xx + c
+ 2 tan x.sec2 x
1
= log (sec2 x + tan2x) + c = 4 tan x sec2x x 2  cos 2 x
4
dt
Q.127
 1 x2
cosec2 xdx =
= tan x sec2x dx
4 (A) cot x + tan–1 x + c (B) cot x – tan–1 x + c
(C) – cot–1 x – tan–1 x + c (D) None of these
2
 1 x  x 2  cos 2 x
Q.108  ex   dx =
1 x2 
Sol.[C]  (1  x 2
) sin 2 x
dx =

ex ex
(A) +c (B)  +c
(1  x 2 ) 2 1 x2 (1  x 2 )  sin 2 x
 (1  x 2 ) sin 2 x
dx
ex
(C) +c (D) None of these
1 x2  1 
  cos ec x  1  x
2
= 2
 dx = – cot x– tan–1x
x x  2 x 
2 
1
Sol.[C] e  (1  x 2 ) 2 
 
dx
+c
x 1 2x 
= e 
 1  x 2  (1  x 2 ) 2

dx

 tan 1 x
1
Q.128  (1  x ) x
dx 
= e . x
c
1 x2 (A) tan–1 x + c (B) (tan–1 x)2 + c
(C) (tan–1 2
x ) +c (D) None of these
1
e tan x
x
Q.109 (log x  (1 / x 2 ))dx 
Sol.[C]  (1  x ) x
dx t = tan –1
(A) ex log x + c (B) ex (log x – 1/x) + c
(C) ex(log x + 1/x) + c (D) (ex /x2) + c x

 1  dt
Sol.[B]  e x  log x  2 dx =  t .2dt dx
=
 x 
1
 log x . e x
x
= dx + ex dx 1 1
I II 2 .
2
1 ( x ) 2 x
1 x
= log x . ex –  x
e dx +
x
1
2
e x dx = t2 + c 2dt =
I II

1 x 1  dx
= log x ex –
x
e  x 2
e x dx 

+ (1  x ) x

1
x 2
e x dx

1 x
= log x ex – e +c
x

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Q.129 If f(x) is the primitive of 2x 2  3  x 1
Q.132 If  (x 2
 1)( x  4)
dx = A log 
2
+
 x 1 
sin 3 x log(1  3x )
x
(tan 1 x ) 2 (e
3
x
 1) B tan–1   , then (A, B) is-
2
(A) (– 1/2, 1/2) (B) (1/2, –1/2)
(x  0), then xlim f ( x ) is-
0 (C) (–1, 1) (D) (1, –1)
(A) 0 (B) 3/5 (2 x 2  3)dx
(C) 5/3 (D) None Sol.[A]  (x 2
 1)( x 2  4)
=

sin x1/ 3 log(1  3x )dx

Sol.[B] f(x) =  (tan 1
x ) 2 (e 5 x
1/ 3
 1)  1  1
lim f ' ( x ) lim
  x 2
1
  dx
2
x 4
x 0
= x 0
1  x 1  1 x
= log   + tan–1 +c
2  x 1 2 2
sin x1 / 3 log(1  3x )
1/ 3
1 1
(tan 1 x ) 2 (e 5 x  1)  a=– , b=
2 2
sin x1 / 3 log(1  3x )
x1 / 3 . .3x
x1 / 3 3x 1 x4
lim
= x 0 2
 tan 1 x   e5 x  1  1 / 3
1/ 3
Q.133 If  xx 5
dx = f(x) + c then
 xx 5
dx =
x.   .5x
 x   5x1 / 3  (A) log |x| – f(x) + c1 (B) log |x| + f(x) + c1
   
(C) x f(x) + c1 (D) None of these
1.1.3
=  3/5 dx
1.1.5 Sol.[A] I1 =  xx 5
= f(x)

( 2 x  sin 2 x )
Q.130  1  cos 2 x
dx =
I2 =
xx
x4
5
dx  I1 + I2 =
1 x 4
xx 5
dx
(A) x tan x + c (B) x cot x + c
1
(C) x(tan x + cot x) + c (D) None of these =
 x dx  I2 = log |x| + c1 – I1

2 x  2 sin x cos x
Sol.[A]  2 cos 2 x
dx
cos 2 x  cos 2
Q.134  cos x  cos 
dx =
 

2
=  x sec x  tan x dx = x tan x + c (A) 2(sin x + x cos ) + c
xf  ( x ) f (x) 
(B) 2(sin x – x cos ) + c
ex
Q.131
 x2
{1 + (x + 2) log (x + 2)} dx = (C) 2 (sin x + 2x cos ) + c
(D) None of these
(A) ex log (x + 2) + c (B) ex /(x + 2) + c cos 2 x  cos 2
(C) ex (x + 2) + c (D) ex (x – 2) + c
Sol.[A]  cos x  cos 
dx

 1  ( 2 cos 2 x  1)  (2 cos 2   1)
Sol.[A]  ex 
 x  2
 log(x  2)  dx = ex log (x + 2)

=
 cos x  cos 
dx

+c

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 cos 2 x  cos 2  (1  nx n –1 – x 2 n )
= 2  cos x  cos 
dx = Q.73  ex
(1 – x n ) 1 – x 2 n
dx =

2   cos x  cos  dx 1– xn 1 x n
(A) ex +c (B) ex +c
= 2(sin x + x cos ) + c 1 xn 1– xn

1– xn 1 x n
(C) – ex + c (D) – ex +c
e 1 xn 1– xn
x
Q.62 [log(sec x  tan x )  sec x ] dx =

(A) ex log tan x + c  (1 – x 2 n )  nx n –1 

x
(B) e log (sec x + tan x) + c Sol.[B]  ex 
 (1 – x n ) 1 – x m 


(C) ex log sec x + c

 1 x n nx n –1 
e   dx
(D) None of these x

 1– xn (1 – x n ) 1 – x 2 n 
Sol.[B] Log (sec x + tan x) = t  
sec x dx = dt 1 x n
f(x) =
 e log (sec x + tan x) + c
x
1– xn
nx n –1
Q.63 The value of  for which f ' (x) =
(1 – x n ) 1 – x 2 n
3 x
4x   4
 4 x x 4
dx = log (4x + x4) is -
  ex
1 x n
+c
1– xn
(A) 1 (B) log e 4
(C) log 4 e (D) 4
Q.74 If f(x) is the primitive of
Sol.[B] 4x + x4 = t
 (4x3 + 4x ln e4) dx = dt
sin 3 x log(1  3x )
dt (x  0), then xlim
  t
= log |t| + c
(tan –1 2
x ) (e
3
x
– 1)
0
f'

   log e 4 (x) is -
(A) 0 (B) 3/5

e log( x 1/ x ) (C) 5/3 (D) None

Q.64  ( x  1) 2 dx is equal to -
sin x1 / 3 log(1  3x )
(A) elog(x+1/x) (B) x
Sol.[D]  f(x) =  (tan –1
x ) 2 (e x
1/ 3
– 1)
(C) log |x| (D) None
 xlim
0 f'(x)
1
1  lim

Sol.[C] x dx x 0
2
x 1
1
 x dx  sin x1/ 3

 x1/ 3
 1/ 3  cos(1  3x ) 
x
   (3x )
   3x 
log |x| + c
2
 tan –1 x   x1 / 3 – 1  1 / 3
  .x. e .x
 x   x 1/ 3 
   
=3

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 = nlim

2x [1 + 2x2 + ……… + nx2n–2]
4
Q.75 Let f(0) = f ' (0) = 0 and f " (x) = sec x + 4, then
 1 1 x 2 
f(x) = = 2x   
2
1 – x (1 – x 2 ) 2 
2 1
(A) log sec x + tan2 x + 2x2 ( given progression is A.G.P.)
3 6
2x
2 1 f(x) =
(B) log cos x + tan2 x + 2x2 (1 – x 2 ) 2
3 6
2x
(C)
2
3
log sec x +
1
6
tan2 x   f ( x )dx =  (1 – x 2 2
)
dx

1 – x2 = t
(D) None of these 2xdx = – dt
Sol.[A] Integrate we have dt
f ' (x) =  sec
2
x.(1  tan 2
x )dx + 4x + C1
=– t 2

1
tan 3 x = +c
f ' (x) = tan x + + 4x + c1 t
3 1
= +c
put x = 0 1– x2
a0
ex
again integrate
Q.81 Let I =
e 4x
 e 2x  1
dx, J =

1 e–x
 tan e
3
f(x) = log sec x + xdx + 2x2 + c2 dx. Then, for an arbitrary
3 – 4x
 e – 2x  1
1 tan 2 x 1 constant C, the value of J – I equals -
= log secx + – log sec x + c2
3 2 3 1  e 4x – e 2x  1 
(A) log  4 x 2x
 +C

+ 2n2 2  e  e 1
put x = 0 1  e 2x  e x  1 
c2 = 0 (B) log  2 x 2x
 +C

2  e – e 1
1 2
2
 f (x )  2x  tan 2 x  log sec x 1  e 2x – e x  1 
6 3 (C) log  2 x x
 +C

2  e  e  1 

 e 4x  e 2x  1 
Q.76 If f(x) = nlim

[2x + 4x3 + ………. + 2nx2n–1], (D)
1
log  4 x  +C
2x 
2  e – e 1
(0 < x < 1) then  f ( x ) dx is equal to - e 3x – e x
1
Sol.[C] J – I =
 1 e 2x
 e 4x
(A) – (1 – x 2 ) (B) Let ex = t
1– x2
 exdx = dt
1 t2 –1
(C)
x2 –1
(D) J-I =
 1 t 2
 t4
dt =

 1
1 t 2 1 – 2 
 t  dt
1– x2   1
t 2  t 2 1 2 
Sol.[D] f(x) = nlim

2 [x+ 2x3 + ……. + nx2n–1], 0 < x <  t 
1

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 1  (log x )
n
Q.100 If In = dx then In + n In–1 is equal
1–
t2
=  1
2 dt to
 t   –1 (A) x (log x)n(B) (x log x)n
 t
(C) (log x) n –1 (D) n(logx) n
1  1 
t+ = u  1 – 2  dt = du
 (log x )
n.
t  t  Sol.[A] In = 1dx

du
= u 2
–1
= (log x ) nx 
 n (log x )
n 1
.
1
x
.x dx
1 u –1
= log
u 1
+c = (log x ) n x  nI n 1
2
1  In + nI n 1 = x (log x ) n
–1ex 
=
1
log
ex +c
2 x 1
e  x 1 Q.65 If  g ( x ) dx = g(x), then
e
e 3x – e x  g ( x ){f ( x )  f ( x )} dx
Sol.[C] J – I =
 1 e 2x
 e 4x is equal to-
Let ex = t (A) g(x) f(x) – g(x) f (x) + c
 exdx = dt
(B) g(x) f (x) + c
t2 –1
J-I =
 1 t 2
 t4
dt = (C) g(x) f(x) + c
(D) g(x) f2 (x) + c

Sol. [C] I =  g( x ) {f ( x ) f ( x )} dx
 1
t 2 1 – 2  using ILATE, we get
 t  dt
  1
I = f(x)  g ( x ) dx –
t 2  t 2 1 2 
 t   (f ( x ).  g ( x ) dx ) dx
1 +  g ( x ) f ( x ) dx
1–
t2 I = f(x) g(x) –  g ( x ) f ( x ) dx
=  1
2 dt
+

 t   –1
t
 g ( x ) f ( x ) dx

I = f(x) g(x) + c
1  1 
t+ = u  1 – 2  dt = du
t  t 
du
= u 2
–1
1 u –1
= log +c
2 u 1

1
x
ex 
–1
1 e
= log +c
2 1
ex  x 1
e

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Q.s

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