Page 236 Integrals CHAPTER 7

CHAPTER 7

I ntegrals

Sol : Delhi 2012

OBJECTIVE QUESTION
x + x2 dx is equal to
7. # 1 +xx++ 1+x
dx
1. The integral # is equal to :
9 - 4x2 (a) 1 1 + x + C (b) 2 (1 + x) 3/2 + C
2 3
(a) 1 sin−1 c 2x m + c (b) 1 sin−1 c 2x m + c (c) 1 + x + C (d) 2 (1 + x) 3/2 + C
6 3 2 3


Sol : OD 2013
(c) sin−1 c 2x m + c (d) 3 sin−1 c 2x m + c
3 2 3


Sol : OD 2024

8. # sin 2x dx is equal to
sin2 x + 2 cos2 x
π/2
2. The value of #π/4 cot θ cosec2 θdθ is (a) − log (1 + sin2 x) + C
(a) 1 (b) - 1 (b) log (1 + cos2 x) + C
2 2
(c) − log (1 + cos2 x) + C
(c) 0 (d) - π (d) log (1 + tan2 x) + C
8


Sol : OD 2024


Sol : SQP 2016

Anti-derivative of tan x − 1 with respect to x is

2 -1
3.
tan x + 1 9. # sec 1(sin
-x
x)
dx 2
is equal to

(b) − sec2 a 4 − x k + c
2 π π
(a) sec a − x k + c (a) sin (tan−1 x) + C (b) tan (sec−1 x) + C
4
(c) log sec a π − x k + c (d) − log sec a π − x k + c (c) tan (sin−1 x) + C (d) − tan (cos−1 x) + C
4 4

Sol :


Sol : OD 2023 Delhi 2010, OD 2008

4. If d
dx f ^x h = 2x + x3 and f ^1 h = 1, then f ^x h is 10. If # 2x x+2 dx = P # 24x + 6 dx + 1 # 2x dx
2
+ 6x + 5 2x + 6x + 5 2 2
+ 6x + 5
(a) x 2 + 3 log | x |+ 1 (b) x 2 + 3 log | x | then the value of P is
(c) 2 - 32 (d) x + 3 log | x |− 4
2
(a) 1 (b) 1


Sol : x OD 2023 3 2
(c) 1 (d) 2
4


Sol : Comp 2007
5. #xe 2 x3
dx equals

(a) 1 ex + C (b) 1 ex + C xe − 1 + ex − 1 dx is equal to

3 4

3 3 11. # xe + ex
(c) 1 ex + C (d) 1 ex + C
3 3

2 2 (a) log (xe + ex ) + C (b) e log (xe + ex ) + C

Sol : OD 2020

(c) 1 log (xe + ex ) + C (d) None of these

e


Sol : OD 2011
6. # 1 + cos x dx is equal to
(a) 2 sin a x k + C (b) 2 sin a x k + C dx
2 2 12. # is equal to
(c) 2 2 sin a x k + C (d) sin a k + C
1 x x (x7 + 1)
2 2 2
CHAPTER 7 Integrals Page 237

# #
7 7 a a
(a) log c 7x m + C (b) 1 log c 7x m + C 19. If f (2a − x) dx = m and f (x) dx = n , then
x +1 7 x +1 0 0
2a
7
(c) log c x +7 1 m + C
7
(d) log c x +7 1 m + C
1 # f (x) dx
0
is equal to
x 7 x


Sol : Comp 2017, Delhi 2008
(a) 2m + n (b) m + 2n
π/2
(c) m - n (d) m + n
13. #0 cos x2 dx is equal to

Sol : Comp 2018

(a) 1 (b) - 2
(c) 2 (d) 0


Sol : Foreign 2009
20. # 1 + cos 2x dx =
(a) 2 cos x + c (b) 2 sin x + c
(c) − cos x − sin x + c (d) 2 sin x + c
3a # b ax - 1 l dx is equal to
1 2
14. 2
0 a-1

Sol : OD 2014, Comp 2012

(a) a − 1 + (a − 1) −2 (b) a + a−2

(d) a2 + 12
x
(c) a - a2


Sol : a SQP 2020
21. # (x xe
+ 1) 2 dx =

(a) ex +c
x
(b) − e + c
(x + 1) 2 x+1
x x
(c) x e+ 1 + c (d) − e 2 + c
(x + 1)


Sol : Foreign 2007

#
a
22. If f (− x) = − f (x) then f (x) dx =
−a
dx is
#
1
15. The value of
ex + e
#
0 a
(a) 2 f (x) dx (b) 0
(a) 1 log b 1 + e l (b) log b 1 + e l 0
e 2 2
(c) 1 (d) - 1
1
(c) log (1 + e) (d) log b 2 l

Sol : Comp 2016
e 1+e


Sol : Foreign 2018, Delhi 2010

β
# #
α
2 23. φ (x) dx + φ (x) dx =
16. The value of #−2 (x cos x + sin x + 1) dx is α β
β
(a) 2 (b) 0 (a) 1 (b) 2 # φ (x) dx
α
(c) - 2 (d) 4
#
α


Sol : Delhi 2015 (c) - 2 φ (x) dx (d) 0
β


Sol : OD 2012, Delhi 2010

17. #0
π/2 sin x − cos x dx is equal to
1 + sin x cos x
# (x) dx =
1
(a) 0 (b) π 24.
4 0

(c) π (d) π (a) 0 (b) 1

2


Sol : OD 2008
(c) 2 (d) 1
2


Sol : Comp 2016

1 + cos 2x dx is equal to
#
π
18.
0 2
dx =
(a) 0 (b) 2 25. # x
(c) 4 (d) - 2


Sol : Foreign 2010, OD 2009 (a) x +k (b) 2 x + k
Page 238 Integrals CHAPTER 7

Sol : SQP 2014

(c) x + k (d) 2 x3/2 + k
3


Sol : OD 2008
33. The anti-derivative of [sec q sec2 q] is
(a) − cot θ + c (b) − cosec θ + c
b
26. #
a
x5 dx = 1
(c) sec2 θ (d) None of these
2
6 6

Sol : Foreign 2017, Delhi 2009
(a) b5 - a5 (b) b - a
6
6 6
(c) a - b (d) a5 - b5
6 34. # cosec x cot xdx is


Sol : Delhi 2007
(a) − cosec x + c (b) - cot x
(c) cosec x (d) None
(tan x) −1 2

Sol : Comp 2008
#
1
27. dx =
0 1 + x2
3
(a) 1 (b) π
2
64 35. # sin x x dx is equal to
(c) π (d) None of these (a) cos x (b) - cos x
192


Sol : Delhi 2010
(c) 2 cos x (d) - 2 cos x


Sol : OD 2007

28. # x dx =
−2
(a) 0 (b) 2
(c) 1 (d) 4


Sol : Foreign 2017

+ p/2
29. #
− p/2
sin9 xdx =
36. # log 2x dx is equal to
(a) - 1 (b) 1 (a) x log x - 1 (b) x log 2x + 1
(c) 0 (d) None of these (c) x log 2x - x (d) x log 2x + 2x


Sol : Delhi 2012, Comp 2009

Sol : Foreign 2018, OD 2013

30. The integration of # 5

x2 $ dx is 37. # sin1 x++sincos2xx dx is equal to
(a) 3 x7 + c (b) 53 ^7 x5 h + c
(a) sin x (b) x
(c) 75 ^5 x7 h + c (d) None of these


Sol : Delhi 2008
(c) cos x (d) tan x


Sol : SQP 2020

31. Which one is equal to # 3

x $ dx ? π/2

(a) 4 4 x3 + c (b) 3 4 x3 + c
38. The value of # log cos x dx
0
is equal to the value of:
3 4 π/2 π/2
(a) # log sin x dx (b) # log sec x dx
(c) 4 3 x 4 + c (d) 3 3 x 4 + c 0 0
3 4 π/2 π/2


Sol : OD 2012
(c) # log cos x dx
0
(d) # log tan x dx
0


Sol : OD 2012

32. # 5e 4 logx
$ dx
5
(a) x 5
(b) x 39. I= # | 1 − x | dx
2
is equal to
5
0
4 5
(c) 5x (d) 4x (a) 0 (b) 1
2
CHAPTER 7 Integrals Page 239

(a) Both (A) and (R) are true and (R) is the correct
(c) 1 (d) None of these explanation of (A).


Sol : Delhi 2007
(b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
#
p
40. sin3 x cos 4 xdx is equal to (c) (A) is true but (R) is false.
-p
(d) (A) is false but (R) is true
(a) 0 (b) 1

Sol :
2 OD 2010
(c) 1 (d) None of these


Sol : Delhi 2018

45. Assertion (A) : # log xdx = x (log x − 1) + k

dx
#
1
41. is equal to Reason (R) :
ex + e−x
0

(a) π (b) tan e - π -1

# (uv) dx = u # vdx − # &dxd (u) # vdx 0 dx .
4 4 (a) Both (A) and (R) are true and (R) is the correct
(c) cot-1 e - π (d) π explanation of (A).
4 2


Sol : OD 2014, Foreign 2008 (b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
dx (c) (A) is true but (R) is false.
42. Assertion (A) : # = tan−1 x + c
a2 − x2 a (d) (A) is false but (R) is true


Sol : SQP 2020, Foreign 2017
Reason (R) : If we let x = a sin θ then given function
becomes constant.
p

(a) Both (A) and (R) are true and (R) is the correct 46. Assertion (A) : #
−p
2
sin7 x dx = 0
explanation of (A). 2

Reason (R) : Here sin7 x is odd function.

(b) Both (A) and (R) are true but (R) is not the
correct explanation of (A). (a) Both (A) and (R) are true and (R) is the correct
explanation of (A).
(c) (A) is true but (R) is false.
(b) Both (A) and (R) are true but (R) is not the
(d) (A) is false but (R) is true correct explanation of (A).


Sol : Comp 2017
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true
4
dx = log 2
43. Assertion (A) : #
2 x


Sol : Delhi 2010

# f ^x hdx = F^b h − F^a h

b
Reason (R) : p

# ` sin x + cos x j dx = 2
2

a 47. Assertion (A) :

− p2
(a) Both (A) and (R) are true and (R) is the correct
# #
a a
explanation of (A). Reason (R) : f (x) dx = 2 f (x) dx ,
−a 0
(b) Both (A) and (R) are true but (R) is not the where f (x) is an even function.
correct explanation of (A). (a) Both (A) and (R) are true and (R) is the correct
(c) (A) is true but (R) is false. explanation of (A).
(d) (A) is false but (R) is true (b) Both (A) and (R) are true but (R) is not the


Sol : Delhi 2018 correct explanation of (A).
(c) (A) is true but (R) is false.
π (d) (A) is false but (R) is true
# sin2 xdx = π
2
44. Assertion (A) :

Sol : Comp 2009
−π 2
2

Reason (R) : Here f ^x h is even function 48. Assertion (A) : Let d {f (x) + c} = F (x) then
dx
if f ^− x h = f ^x h : # f ^x h dx = 2 # f ^x h dx .
a a

−a 0
# F (x) dx = f (x) + c
Reason (R) : Integration is inverse process of
Page 240 Integrals CHAPTER 7

differentiation. 52. If n 2 1, then

! 1
(a) Both (A) and (R) are true and (R) is the correct
explanation of (A). Assertion (A) : # 1 +dxx n = # (1 −dxx )
n 1/n
0 0
(b) Both (A) and (R) are true but (R) is not the b b
correct explanation of (A). Reason (R) : # f (x) dx = # f (a + b + x) dx
(c) (A) is true but (R) is false. a a

(d) (A) is false but (R) is true (a) Both (A) and (R) are true and (R) is the correct


Sol : Foreign 2009 explanation of (A).
(b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
49. Assertion (A) : # xex dx = ex + C
x+1 (c) (A) is true but (R) is false.
(x + 1) 2
(d) (A) is false but (R) is true
Reason (R) : # ex {f (x) + f ' (x)} dx = ex f (x) + C

Sol : Comp 2015, OD 2011
(a) Both (A) and (R) are true and (R) is the correct
explanation of (A).
2π
(b) Both (A) and (R) are true but (R) is not the 53. Assertion (A) : # sin3 x dx = 0
correct explanation of (A). 0

(c) (A) is true but (R) is false. Reason (R) : sin3 x is an odd function
(d) (A) is false but (R) is true (a) Both (A) and (R) are true and (R) is the correct


Sol : Delhi 2007 explanation of (A).
(b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
50. Assertion (A) : # ex + dx
e−x + 2
= 1 +C
ex + 1 (c) (A) is true but (R) is false.
d {f (x)} (d) (A) is false but (R) is true
Reason (R) : # =− 1 + C

Sol : SQP 2016
{f (x)} 2 f (x)
(a) Both (A) and (R) are true and (R) is the correct
explanation of (A).
(b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true


Sol : OD 2018

54. Assertion (A) :

# sin 3x cos 5x dx = − cos 8x + cos 2x + C

π/2
51. Assertion (A) : # sin x dx = 2 16 4
− π/2
Reason (R) : 2 cos A sin B = sin (A + B) − sin (A − B)
Reason (R) : (a) Assertion is true, Reason is true; Reason is a
b c b correct explanation for Assertion.
# f (x) dx = # f (x) dx + # f (x) dx, where c ! (a, b)
a a c (b) Assertion is true, Reason is true; Reason is not a
correct explanation for Assertion.
(a) Both (A) and (R) are true and (R) is the correct (c) Assertion is true; Reason is false.
explanation of (A). (d) Assertion is false; Reason is true.
(b) Both (A) and (R) are true but (R) is not the

Sol : OD 2017
correct explanation of (A).
(c) (A) is true but (R) is false. π

tan x dx = π
#0
2
(d) (A) is false but (R) is true 55. Assertion (A) : I =


Sol : OD 2012
2
2
Reason (R) : tan x = t makes the integrand in I as a
rational function.
CHAPTER 7 Integrals Page 241

(a) Both (A) and (R) are true and (R) is the correct 67. Evaluate # dx .
explanation of (A).

Sol : 1 - x2 OD 2011, Delhi 2008
(b) Both (A) and (R) are true but (R) is not the
correct explanation of (A). −1

(c) (A) is true but (R) is false. 68. Evaluate # e tan x dx .

1 + x2
(d) (A) is false but (R) is true

Sol : OD 2011

Sol : Foreign 2013

69. Evaluate # x3 − x2 + x − 1 dx .


Sol : x−1 Comp 2011


VERY SHORT Answer QUESTION

70. Write the value of # 1 - sin

2
x dx .
cos x


Sol : Foreign 2011
56. Find : # cos3 xelogsinx dx


Sol : OD 2024

71. Evaluate # x3 - 1 dx .
x2
57. Find : # 5 + 41x − x2 dx

Sol : Delhi 2010,


Sol : OD 2024

72. Evaluate # sec2 (7 - 4x) dx .

1
58. Evaluate #-1 x 4 - x dx .

Sol : Delhi 2010; OD 2010


Sol : OD 2023

73. Evaluate # dx .
sin2 x - cos2 x dx .

Sol : (x2 + 1) (x2 + 2) Foreign 2010
59. Find #
sin2 x cos2 x


Sol : Foreign 2014

log x
74. Evaluate # x
dx .
6
sin x dx .

Sol : Foreign 2010, OD 2008
60. Find #
cos8 x


Sol : Foreign 2014

75. Evaluate # 2x dx .


Sol : Foreign 2010
61. Evaluate # cos-1 (sin x) dx .


Sol : Foreign 2014

76. Evaluate # cos 2x + 2 sin2 x dx .

Sol : cos2 x OD 2018
62. Evaluate # (1 - x) x dx .


Sol : Delhi 2012, Comp 2010

77. Find # dx .
2

Sol : x2 + 4x + 8 Delhi 2017
63. Evaluate # 1 + cos 2x
dx .


Sol : Foreign 2012

78. Find : # 3 - 5 sin x dx .

Sol : cos2 x Foreign 2018
64. Write the value of # dx .
x2 + 16


Sol : Foreign 2011
3
79. Evaluate #2 3x dx .


Sol : Delhi 2017
65. Write the value of # 2 - 3 sin x dx .


Sol : cos2 x Delhi 2011
π/4
80. Evaluate #0 tan x dx .


Sol : Foreign 2014, Delhi 2013
66. Write the value of # sec x (sec x + tan x) dx .


Sol : Delhi 2011
Page 242 Integrals CHAPTER 7

# ^x +21xh2+^x1− 1h dx
1
#0
2
81. Evaluate x ex dx . 94. Find :


Sol : Foreign 2014


Sol : OD 2024

π/4
82. Evaluate #0 sin 2x dx . log 3 1


Sol : Foreign 2014 95. Evaluate #log 2 ^ex + e−x h^ex − e−x h
dx


Sol : OD 2023

83. Evaluate #0
1 1 dx .


Sol : 1 - x2 Comp 2014, OD 2011 sin-1 x
96. Find #


Sol :
^1 - x2h3/2 OD 2023
3
84. Evaluate #
2 x - 1 dx .
x2


Sol : # ex b 11 -- cos xl
sin x dx
1 Comp 2014
97. Evaluate


Sol : OD 2023

85. Evaluate #2
3 1 dx .


Sol : x 4
Delhi 2012
98. Find the value of #1 x - 5 dx .


Sol : OD 2020

86. Evaluate #0
1 1 dx .
;x1 - 1 2 E e2x dx .


Sol : 1 + x2 2
Comp 2011
99. Evaluate #1 2x


Sol : OD 2020

2
87. Evaluate #0 4 - x2 dx .


Sol : OD 2012

88. Evaluate #
3 dx .


Sol : 1 1 + x2 Foreign 2011, Delhi 2009

π/2 x
89. #0 e (sin x − cos x) dx .
x ^1 - x hn dx
Evaluate 1


Sol : Delhi 2014 100. Find the value of #0


Sol : OD 2020

90. Evaluate #
e2 dx .


Sol : e x log x OD 2014 101. Find # sin2 x - cos2 x dx .


Sol : sin x cos x OD 2017, Delhi 2010

91. Write the value of #0

1 ex dx .
1 + e2x 102. Evaluate # dx .


Sol : Comp 2012
sin2 x cos2 x


Sol : Delhi 2014C; Foreign 2014

92. Evaluate #0
1 2x dx .


Sol : 1 + x2 OD 2011C, 2008 103. Write the anti-derivative of c 3 x + 1 m .


Sol : x Delhi 2014

SHORT ANSWER QUESTION 104. Given, # ex (tan x + 1) sec x dx = ex f (x) + C . Write

f (x) satisfying above.


Sol : OD 2012; Foerign 2011

π e cos x
93. Evaluate : #0 e cos x
+ e− cos x
dx
x + cos 6x dx .


Sol : OD 2024 105. Write the value of #
3x2 + sin 6x


Sol : Comp 2012
CHAPTER 7 Integrals Page 243

106. Write the value of # sec2 x dx . 120. Find # sin3 x + cos3 x dx .

cosec2 x

Sol : sin2 x cos2 x OD 2019, Delhi 2017


Sol : Delhi 2012C, 2011

(log x) 2 121. Find # x - 3 ex dx .

107. Evaluate # x
dx .


Sol :
(x - 1) 3
OD 2019


Sol : OD 2011

122. Find # x - 5 ex dx .
108. Evaluate # (ax - b) 3 dx .

Sol : (x - 3) 3 OD 2019


Sol : OD 2011 , Foreign 2007

123. Find # dx .
(1 + log x) 2

Sol : 5 - 8x - x2 OD 2017
109. Evaluate # x
dx .


Sol : Foreign 2011; Delhi 2009

124. Find # x dx .
2x −2x


Sol : a - x3
3
Delhi 2016, OD 2012
110. Evaluate # e2x − e−2x dx .
e +e


Sol : Foreign 2011

125. Evaluate # x sin-1 x dx .

Sol : 1 - x2 Foreign 2016; Delhi 2012

111. Evaluate # cos x dx .

Sol : x Foreign 2011 sin (x − a)
126. Evaluate # sin (x + a)
dx .


Sol : Foreign 2015; Delhi 2013

112. Evaluate # 2 cos x dx .

3 sin2 x (x2 + 1) ex


Sol : Comp 2011
127. Find # dx .


Sol : (x + 1) 2 Delhi 2014

113. Evaluate # 2 cos x dx .

Sol : sin2 x OD 2011C, 2009, 2009 x3
128. Find # dx .


Sol : x + 3x 2 + 2
4
Comp 2014

114. Find : # sec2 x dx .

Sol : tan2 x + 4 Delhi 2019 x cos-1 x dx .
129. Evaluate #


Sol : 1 - x2 OD 2014C; Foreign 2014

115. Find : # 1 - sin 2x dx , p 1 x 1 p .

Sol : 4 2 Delhi 2019, Comp 2010
130. Evaluate # sin6 x + cos6 x dx .


Sol : sin2 x cos2 x Comp 2014, Delhi 2010

116. Find : # sin-1 (2x) dx .

Sol : Delhi 2019
131. Evaluate # cos 2x − cos 2α dx .


Sol : cos x − cos α OD 2013

117. Find the value of # tan2 x $ sec2 x dx .

Sol : 1 - tan6 x Delhi 2019
132. Evaluate # dx .


Sol : x (x5 + 3) OD 2013

118. Find the values of # sin x $ log cos x dx .

Sol : Delhi 2019
133. Evaluate # dx .


Sol : x (x3 + 1) OD 2013

119. Find # 3 - 2x - x2 dx .


Sol : OD 2019
Page 244 Integrals CHAPTER 7

134. Evaluate # dx . 2 x


Sol : x (x3 + 8) OD 2013
148. Evaluate #-1 x
dx .


Sol : Delhi 2019

135. Evaluate # sin x $ sin 2x $ sin 3x dx . 149. Evaluate #-1

2
x3 - x dx .


Sol : Delhi 2012, Comp 2009


Sol : Delhi 2016; OD 2010

136. Evaluate # 1 − sin x e −2x dx . π/4

Sol : 1 + cos x Comp 2013

150. Evaluate #0 log (1 + tan x) dx .


Sol : Comp 2015, OD 2010, Comp 2013

+ sin x ex dx .
# b 11 + cos x l
2
137. Evaluate #−2 1 +x 5x dx .
2
151. Evaluate


Sol : Comp 2012


Sol : OD 2016

138. Evaluate # x2 + 4 dx . π/2 2

Sol : x 4 + 16 Comp 2011, Foreign 2008

152. Evaluate #0 x sin x dx .


Sol : Comp 2014

139. Evaluate # 1 - x2 dx . 153. Evaluate #2

5
[ x − 2 + x − 3 + x − 5 ] dx .


Sol : x (1 - 2x) Delhi 2010

Sol : Delhi 2013

140. If #0
a 1 dx = π , then find the value of a .
4 + x2 8 2π 1
154. Evaluate #0 dx .


Sol : 1 + e sinx OD 2013


Sol : OD 2014, Delhi 2010

5
155. Evaluate #2 [ x + x − 2 + x − 4 ] dx .
141. If f (x) = #0
x
t sin t dt , then write the value of f l (x).

Sol : Delhi 2013


Sol : OD 2014

3
156. Evaluate # [ x − 1 + x − 2 + x − 3 ] dx .
4 dx .

Sol : 1 Delhi 2013
142. Evaluate #2
9 + x2


Sol : OD 2014
157. Evaluate #0
π x tan x dx .


Sol : sec x $ cosec x Delhi 2011C, 2008, OD 2009

143. Evaluate #0
3 dx .


Sol : 9 + x2 Delhi 2014
158. Evaluate #π/6
π/3 dx .


Sol : 1 + tan x OD 2011
−1
144. Evaluate #0
1 tan x dx .


Sol : 1 + x2 Comp 2014, OD 2009
- sin 2x dx .
# ex b 11 - cos 2x l
159. Evaluate


Sol : Comp 2013
π/4
145. Evaluate #− π/4 sin3 x dx .


Sol : Comp 2010
sin x - cos x dx .
160. Evaluate #


Sol : sin 2x Comp 2011, OD 2007
π/2
146. Write the value of the following integral #− π/2 sin5 x dx .


Sol : OD 2010
161. Evaluate # x2 + 1 dx .


Sol : x4 + 1 Comp 2011
π
147. Evaluate #− π (1 − x ) sin x $ cos
2 2
x dx .


Sol : Delhi 2019
CHAPTER 7 Integrals Page 245

# ex b 1sin 4x - 4 dx . cos θ
- cos 4x l
162. Evaluate 173. Find # dθ .


Sol : Delhi 2010

Sol : (4 + sin2 θ) (5 − 4 cos2 θ) OD 2017, Comp 2015

163. Evaluate #
4
( x − 1 + x − 2 + x − 4 ) dx . (3 sin θ − 2) cos θ


Sol :
174. Find # dθ .
1 OD 2017, Comp 2011 5 − cos2 θ − 4 sin θ


Sol : Delhi 2016

LONG ANSWER QUESTION 175. Find # (x + 3) (3 − 4x − x2) dx .

Sol : OD 2016, Delhi 2015, Comp 2014

176. Evaluate # x2 + x + 1 dx .
x
164. Find # x 2 + 3x + 2
dx

Sol :
2
(x + 1) (x + 2) OD 2016, Delhi 2015, Comp 2009


Sol : OD 2020

3x + 5 dx . (2x - 5) e2x
165. Find : # 177. Find # dx .


Sol : x2 + 3x − 18 Delhi 2019

Sol : (2x - 3) 3 OD 2016

178. Find # (2x + 5) 10 − 4x − 3x2 dx .

Sol : Foreign 2016

(x2 + 1) (x2 + 4)
179. Find # dx .


Sol : (x2 + 3) (x2 − 5) Foreign 2016, OD 2007

166. Find the value of # (1 + sincos x

x) (2 + sin x)
dx . dx


Sol : Delhi 2019 180. Find # sin x + sin 2x
.


Sol : Delhi 2015

167. Find # x2 + x + 1 dx . 2


Sol : (x + 2) (x2 + 1) OD 2019, Delhi 2008 181. Integrate w.r.t. x , x − 3x +2 1 .


Sol : 1−x Delhi 2015

168. Find # 2 cos x dx .

Sol : (1 - sin x) (2 - cos2 x) OD 2019
182. Evaluate # (3 − 2x) 2 + x − x2 dx .


Sol : OD 2015

169. Find # 2 cos x dx . log x

Sol : (1 − sin x) (1 + sin2 x) OD 2018, Foreign 2007 183. Find # dx .


Sol : (x + 1) 2 OD 2015, Delhi 2010

170. Find # 4 dx .


Sol : (x − 2) (x2 + 4) Comp 2018 184. # e2x $ sin (3x + 1) dx .


Sol : Foreign 2015

171. Find # 2x dx .
(x2 + 1) (x2 + 2) 2 185. Evaluate # x2 dx .


Sol : Delhi 2017
(x2 + 4) (x2 + 9)


Sol : Foreign 2015; Delhi 2013

172. Find # 2x dx .


Sol : (x2 + 1) (x 4 + 4) Delhi 2017
186. Evaluate # (x − 3) x2 + 3x − 18 dx .


Sol : Delhi 2014
Page 246 Integrals CHAPTER 7

x+2 # e2x b 11++sin

cos x l
187. Evaluate # dx . 201. Evaluate 2x dx .


Sol : x 2 + 5x + 6 OD 2014

Sol : Comp 2010

# ;log (log x) +
(log x) 2 E
188. Evaluate # (3x − 2) x2 + x + 1 dx . 202. Evaluate 1 dx .


Sol : Foreign 2014

Sol : Comp 2010

189. Find # 5x − 2 dx . x+2

Sol : 1 + 2x + 3x2 Delhi 2014C; Delhi 2013
203. Evaluate # dx .


Sol : (x − 2) (x − 3) OD 2010

190. Evaluate # 3x + 1 dx . 1


Sol : (x + 1) 2 (x + 3) 204. Evaluate # dx .
Comp 2013

Sol : sin 4 x + sin2 x cos2 x + cos 4 x OD 2014, Delhi 2010

191. Evaluate # 2x2 + 1 dx . 205. Evaluate #^ cot x + tan x h dx .

Sol : x (x2 + 4)
2
Delhi 2013

Sol : OD 2014, Delhi 2010

x2 + 1 206. Evaluate # 1 dx .
192. Evaluate # dx .

Sol : cos 4 x + sin 4 x Delhi 2014, Comp 2010


Sol : (x + 4) (x2 + 25)
2
Delhi 2013, SQP 2012

193. Evaluate # x+2 dx .

Sol : x 2 + 2x + 3 OD 2013

194. Evaluate # 3x + 5 dx .


Sol : x3 − x2 − x + 1 Comp 2013, OD 2011

207. Find # x2 dx .
2

Sol : (x + 1) (x2 + 4)
2
Delhi 2014
195. Evaluate # dx .


Sol : (1 − x) (1 + x2) Delhi 2012

208. Find # sin−1 x − cos−1 x dx , x d [0, 1].

x 2


Sol : sin−1 x + cos−1 x OD 2014
196. Evaluate # dx .
(x sin x + cos x) 2


Sol : Comp 2012
209. Find # x2 + x + 1 dx .


Sol : (x + 1) 2 (x + 2) Comp 2014, OD 2010

197. Evaluate #e 2x
sin x dx .


Sol : Foreign 2011

x2 + 1 ` log x2 + 1 − 2 log x j
210. Find # dx .
x4
198. Evaluate # 3x + 5 dx .


Sol : x 2 − 8x + 7 Foreign 2011, Delhi 2008

Sol : Foreign 2014

2x 211. # x2 + 1
199. Evaluate # dx . Evaluate
(x − 1) 2 (x + 3)
dx .
(x2 + 1) (x2 + 3)

Sol : Delhi 2012

Sol : Delhi 2011

212. Evaluate # 6x + 7 dx .
5x + 3

Sol : (x − 5) (x − 4) OD 2011, Foreign 2009
200. Evaluate # 2
dx .


Sol : x + 4x + 10 Delhi 2011; OD 2010
CHAPTER 7 Integrals Page 247

a a
213. Prove that #0 f (x) dx = #0 f (a − x) dx , hence evaluate 226. Evaluate #
2 5x 2 dx .
2


Sol : 1 x + 4x + 3 OD 2011
I = # x sin x2 dx .
π

0 1 + cos x


Sol : Delhi 2019

1 log 1 + x
227. Evaluate #0 dx .
a a

Sol : 1 + x2 Foreign 2011
214. Prove that
0
#0
f (x) dx = # f (a − x) dx . and hence
evaluate #
π/2 x dx .
0 sin x + cos x
log 1 - 1 dx .
1


Sol : OD 2019, Delhi 2007 228. Evaluate #0


Sol : x OD 2011

215. Evaluate #0
π x sin x dx .
1 + cos2 x 229. Evaluate #0
π x dx .


Sol : Delhi 2017; OD 2013, 2012, 2011C, 2008 1 + sin x


Sol : Delhi 2010

#0
π x tan x dx . π/4 sin x + cos x dx .
216. Evaluate
sec x + tan x 230. Evaluate #0 16 + 9 sin 2x


Sol : OD 2017

Sol : OD 2018

e $ sin a π + x k dx . sin x + cos x dx .

π 2x π/4
217. Evaluate #0 4 231. Evaluate #0


Sol : Delhi 2016

Sol : 9 + 16 sin 2x Foreign 2014; Delhi 2014, 2011

#0
π x π x
218. Evaluate
1 + sin α sin x
dx . 232. Evaluate #0 dx .


Sol : Foreign 2016, OD 2014

Sol : a2 cos2 x + b2 sin2 x Foreign 2014; OD 2009

π
219. Evaluate #− π (cos ax − sin bx) 2 dx . 233. Evaluate #0
π x tan x dx .


Sol : Delhi 2015

Sol : sec x + tan xForeign 2014; Delhi 2014C, 2010, 2008; OD 2008

220. Find #0
π/4 dx . π/3 dx


Sol : cos3 x 2 sin 2x OD 2015
234. Evaluate #π/6 .


Sol : 1 + cot x Delhi 2014

#− π/2 1cos x dx .
π/2
2 $ π.
π/4
221. Evaluate 235. Prove that #0 ( tan x + cot x ) dx =


Sol : + ex Foreign 2015

Sol : 2 Delhi 2012

222. #π/6
π/3 sin x + cos x dx . p/2
Evaluate
sin 2x
236. Evaluate #p/4 cos 2x $ log (sin x) dx .


Sol : OD 2014C; Delhi 2011

Sol : Foreign 2012

223. Prove that #0

π/2 sin2 x dx = 1 log ( 2 + 1). 237. Evaluate #0
π/2
2 sin x cos x tan−1 (sin x) dx .


Sol : sin x + cos x 2 Foreign 2014

Sol : Delhi 2011

#0
1 x 4 + 1 dx .
224. Evaluate


Sol : x2 + 1 Foreign 2011

Case Based Question

225. Evaluate #0
π/2 x + sin x dx .
1 + cos x 238. Commodity prices are primarily determined by the


Sol : OD 2011, Comp 2008
forces of supply and demand in the market. For
example, if the supply of oil increases, the price of
one barrel decreases. Conversely, if demand for oil
Page 248 Integrals CHAPTER 7

increases (which often happens during the summer), any time t . Initially ^t = 0h 50 g of the first substance
the price rises. Gasoline and natural gas fall into the was present; 1 hr later, only 10 g of it remained.
energy commodities category. (i) Find an expression that gives the amount of the
first substance present at any time t .
(ii) What is the amount present after 2 hr?


Sol :

240. Chemical reaction, a process in which one or more

substances, the reactants, are converted to one or more
different substances, the products. Substances are
either chemical elements or compounds. A chemical
reaction rearranges the constituent atoms of the
reactants to create different substances as products.

The price p (dollars) of each unit of a particular

commodity is estimated to be changing at the rate
dp
= − 135x2
dx 9+x
where x (hundred) units is the consumer demand (the
number of units purchased at that price). Suppose 400
units ^x = 4h are demanded when the price is $30 per
unit.
(i) Find the demand function p ^x h .
(ii) At what price will 300 units be demanded? At
what price will no units be demanded? In a certain chemical reaction, a substance is converted
(iii) How many units are demanded at a price of $20 into another substance at a rate proportional to the
per unit? square of the amount of the first substance present at
any time t . Initially ^t = 0h 50 g of the first substance


Sol :
was present; 1 hr later, only 10 g of it remained.
239. Chemical reaction, a process in which one or more (i) Find an expression that gives the amount of the
substances, the reactants, are converted to one or more first substance present at any time t .
different substances, the products. Substances are (ii) What is the amount present after 2 hr?


Sol :
either chemical elements or compounds. A chemical
reaction rearranges the constituent atoms of the
reactants to create different substances as products. 241. Bata India is the largest retailer and leading
manufacturer of footwear in India and is a part of the
Bata Shoe Organization. Incorporated as Bata Shoe
Company Private Limited in 1931, the company was
set up initially as a small operation in Konnagar (near
Calcutta) in 1932. In January 1934,

In a certain chemical reaction, a substance is converted

into another substance at a rate proportional to the
square of the amount of the first substance present at
CHAPTER 7 Integrals Page 249

(i) How many items can Bob memorize during the

first 10 minutes?
(ii) How many additional items can he memorize
during the next 10 minutes (from time t = 10 to
t = 20 )?


Sol :

243. Different types of drugs affect your body in different

ways, and the effects associated with drugs can vary
from person to person. How a drug effects an individual
is dependent on a variety of factors including body
size, general health, the amount and strength of the
The manager of BATA show room at Jaipur determines drug, and whether any other drugs are in the system
that the price p (dollars) for each pair of a popular at the same time. It is important to remember that
brand of sports sneakers is changing at the rate of illegal drugs are not controlled substances, and
pl^x h = −2 300x3/2
therefore the quality and strength may differ from one
^x + 9h batch to another.
when x (hundred) pairs are demanded by consumers.
When the price is $75 per pair, 400 pairs ^x = 4h are
demanded by consumers.
(i) Find the demand (price) function p ^x h .
(ii) At what price will 500 pairs of sneakers be
demanded? At what price will no sneakers be
demanded?
(iii) How many pairs will be demanded at a price of
$90 per pair?


Sol :

The concentration C ^ t h in milligrams per cubic

242. Bob is taking a learning test in which the time he centimeter ^mg/cm3h of a drug in a patient’s
takes to memorize items from a given list is recorded. bloodstream is 0.5 mg/cm3 immediately after an
Let M ^ t h be the number of items he can memorize in injection and t minutes later is decreasing at the rate
t minutes. His learning rate is found to be
C l^ t h = − 00.01
0.01t
.01e 3
M l^ t h = 0.4t − 0.005t 2
^e t + 1h
2 mg/cm per minute

A new injection is given when the concentration drops

below0.05 mg/cm3 .
(i) Find an expression for C ^ t h .
(ii) What is the concentration after 1 hour? After 3
hours?


Sol :

244. In mathematics, a continuous function is a function

such that a continuous variation of the argument
induces a continuous variation of the value of the
function. This means that there are no abrupt changes
Page 250 Integrals CHAPTER 7

in value, known as discontinuities.

Let f (x) be a continuous function defined on [a, b],

then
b b
# f (x) dx = # f (a + b − x) dx

a 0

On the basis of above information, answer the

following questions.
(i) Evaluate #
2 x dx
1 x + 3−x
p

#p
3
(ii) Evaluate log tan x dx
6
1
b xn
iii) Evaluate #a 1 1 dx ]
x + (a + b − x) n
n

b f (x)
(iv) Evaluate # dx
a f (x) + f (a + b − x)

Sol :

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