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MATHEMATICS: QUESTION BANK
CHAPTER 7: INTEGRALS(INDEFINITE)

Standard forms
1mark questions:
Write an antiderivative for each of the
following functions using differentiation
Question 1: i)sin 2x
Soln: The anti derivative of sin 2x is a
function of x whose derivative is
sin 2x
...

It is known that,

Question 6:

Therefore, the anti derivative of

Question 7: Find an anti-derivative
of
with respect to x
...

Question 3: e2x
The anti derivative of e2x is the function of x
whose derivative is e2x
...

√
√
) = sinx+cosx
√(
Antiderivative of sinx+cosx is
cosx-sinx+c

Therefore, the anti derivative of


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)


...


Therefore, the anti derivative of

...

It is known that,

Therefore, the anti derivative of
is


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Question 8:

Question 12:

Question 13:

Question 9:

Question 14:

Question 10:

Question 15:Find the anti derivative of

Solution:

Question 11:

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Evaluate  tan 2 (2 x)
...

2

2
...
dx   (sec 2 2 x  1)dx

= -2cot +c

Solution:  1 tan 2 x  x  c

2

TWO MARK QUESTIONS:

Question 3:

Integrate the following w
...
t x
Let 1 + log x = t
1
...

Hint: log |x| = t

∴

Question 4:sin x ⋅ sin (cos x)
sin x ⋅ sin (cos x)
Let cos x = t
∴ −sin x dx = dt

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Question 17:

Question 13:
Let

Let sin 2x = t ∴

∴ 2xdx = dt

Question 18:
Let
∴ cos x dx = dt

Question 14:
Let

∴ 2xdx = dt

Question 19: cot x log sin x
Let log sin x = t

Question 15:
∴

Let

Question 20:
Let 1 + cos x = t ∴ −sin x dx = dt
Question 16:
Let

∴

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∴ dx = dt

Question 3:
Let
∴ dx = dt

Question 6:
∴

Let

Question 7:
Question 4:
Let

Let 2x − 3 = t ∴ 2dx = dt

∴

Question 8:
Let 7 − 4x = t ∴ −4dx = dt

Question 5:
Dividing numerator and denominator by ex,
we obtain

Question 9:
Let
Let

∴

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Question 15:

Question 17:
Let x4 = t
∴ 4x3 dx = dt

Let
∴
From (1), we obtain

Question 16:

Let
∴

******

INTEGRATION USING TRIGONOMETRIC IDENTITIES:
THREE MARKS QUESTIONS:
Integrate the following functions:
Question 2:
It is known that,
Question 1:

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Question 9:

Question 12:

Question 10: sin4 x

Question 13:

Question 14:
4

Question 11: cos 2x

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Question 21: sin−1 (cos x)
Question 22:

Question 23:

Question 24:
It is known that,
Let exx = t

Substituting in equation (1), we obtain

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Question 9:
Let tan x = t ∴ sec2x dx = dt

Question 12:

Question 10:

Question 13:
Question 11:

Question 14:

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From (1), we obtain

Question 6: Integrate

x2
x  2x  3
2

with respect to x
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Substituting equations (2) and (3) in equation
(1), we obtain
Substituting equations (2) and (3) in (1), we
obtain

Question 3:

Question 2:

Equating the coefficients of x and constant
term on both sides, we obtain
Equating the coefficients of x and constant
term, we obtain 2A = 6 ⇒ A = 3
−9A + B = 7 ⇒ B = 34
∴ 6x + 7 = 3 (2x − 9) + 34

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Using equations (2) and (3) in (1), we obtain

INTEGRATION BY PARTIAL FRACTIONS
TWO MARK QUESTIONS:
Question 1:
Let
Equating the coefficients of x and constant
term, we obtain
A + B = 1 ; 2A + B = 0
On solving, we obtain A = −1 and B = 2

THREE MARK QUESTIONS:
Question 1:
Let

Substituting x = 1, 2, and 3 respectively in
equation (1), we obtain
A = 1, B = −5, and C = 4

Question 2:
Let
Equating the coefficients of x and constant
term, we obtain
A + B = 0 ; −3A + 3B = 1
On solving, we obtain

Question 2:
Let

Substituting x = 1, 2, and 3 respectively in
equation (1), we obtain

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Therefore, on dividing (1 −
x2) by x(1 − 2x), we obtain

Let

Question 5:

Substituting x = 0 and in equation (1), we
obtain A = 2 and B = 3

Substituting in equation (1), we obtain

Let

Substituting x = 1, we obtain
Equating the coefficients of x2 and constant
term, we obtain
A + C = 0 ;−2A + 2B + C = 0
On solving, we obtain

Question 6:

Question 4:
Let
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Therefore, on dividing (x3 + x + 1) by x2 − 1,

On solving, we obtain

we obtain
Let
Substituting x = 1 and −1 in equation (1), we
Question 7:

obtain

Let

Equating the coefficients of x2 and x, we
obtain

Question 10:

Equating the coefficient of x and constant
term, we obtain
A=3
2A + B = −1 ⇒ B = −7

Question 8:

Let

Substituting x = −1, −2, and 2 respectively in
equation (1), we obtain
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Substituting x = 1 and 2 in (1), we obtain
A = −1 and B = 2

Question 15:
[Hint: Put ex = t]
Let ex = t ⇒ ex dx = dt

Question 17:
Substituting t = 1 and t = 0 in equation (1), we
obtain A = −1 and B = 1

Equating the coefficients of x2, x, and constant
term, we obtain
A + B = 0; C = 0 ; A = 1
On solving these equations, we obtain
A = 1, B = −1, and C = 0

Question 16:

ADDITIONAL QUESTIONS: 4 TO 5 MARKS:
Question 1:

Equating the coefficient of x2, x, and constant
term, we obtain
A − B = 0; B − C = 0; A + C = 2
On solving these equations, we obtain
A = 1, B = 1, and C = 1

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Given Integral

Let I =
Taking x as first function and sin 3x as second
function and integrating by parts, we obtain

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=

∫

=

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Question 4: x logx
Let
Taking log x as first function and x as second
function and integrating by parts, we obtain

THREE MARKS QUESTIONS:
Integrate the following w
...
t
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Question 3:
Let
Taking
as first function and x as
second function and integrating by parts, we
obtain

Question 4: Evaluate:

∫

∫

∫
∫

=


...


∫

∫
∫

=

|

|

Question 5:
Let
Taking cos−1 x as first function and x as second
function and integrating by parts, we obtain

Question 6:
Let
Taking
as first function and 1 as
second function and integrating by parts, we
obtain

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Evaluate:

e

x

 1  sin x 

 dx
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Let

⇒

Let
⇒
It is known that,

It is known that,

Question 3:

Question 2:

Let
Integrating by parts, we obtain

Again integrating by parts, we obtain

Let
⇒
It is known that,

Question 4:
Let

⇒

From equation (1), we obtain
= 2θ
⇒
Question 3:
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Find the integral of

with

√

respect to x and hence evaluate
∫


...
Find the integral of

with respect

to x and evaluate ∫


...
Find the integral of

∫

Consider

(

(
[

)(

)
(

)

(
)(

)
[

with

√

respect to x and hence evaluate
∫


...


|

)
]

(

]
)

(

]
)

*(

∫
|

)

|

+
|]

|
∫

(

Let

)

then

)

)

∴

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4
...
Find the integral of

+

|


...
Find the integral of



(

=

]
)

=∫

√

Let I= 

∫

=

=x2+2x+1=(x2+2x+1)+1

∫

)
]

|

Consider

∴

√

=(x+1)2+(1)2

[

∴


...
Find the integral of √

=∫

respect to x and evaluate ∫ √

=log|sec +tan |+C1
=log|
= log|
=log|

√
√
√

with

Solution: Let
∫√
integration by parts

|+C1
|-log|a|+C1
|+C

We get
√

where C=C1 –log|a|

√
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√

|

√

√

|

|

√

Consider ∫ √

|

√

Now consider

I=∫[(

|
with

Solution: Let
∫√
integration by parts

√

applying

√

√

√

( )

√
Consider ∫ √

(

)

√

)
(

)

(

Solution: Let
∫√
integration by parts

√

)

=

applying

√

∫√
(

)

√

∫

√

∫ *√

√

√

√

+

∫√
√

|

√

√

Find the integral of √
with
respect to x and hence evaluate
∫

We get

|

|

Note: In this chapter “Indefinite Integrals”
Some of the solved examples given in the text
book are not included in the question bank
...


( )

(

=

+

Note: The above questions is for 5 mark
questions in part D of the question paper for
second PUC
...


We get

√

Put x+1=t dx=dt

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Assignments
(i) Integration by substitution
LEVEL I
1

sec 2 (log x )
dx
1
...

dx
1 x2

3
...




1
dx
x x

2
...


1

 e x  1 dx

LEVEL III
1
...
cos x dx

2
...


1

 sin x
...


 sin

3

2
2
...
dx

x
...


 cos x
...
cos 3x
...


 9x 2  12x  13

3
...




LEVEL II
1
...
tan x
...




sin 4x
dx
sin x
LEVEL III

1
...
dx

2
...
cos 3 x
...




dx

2
...


x

 x 4  x 2  1 dx

2
...




4
...


1 x
dx
1 x

x2  x 1
 x 2  x  1 dx

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x2
x  5x  6
2

dx

1
...


x2
dx
( x  1)( x  2)( x  3)

3
...


x2  x 1

2

 x 2 (x  2)

2
...


 (x  1) 2 (x  3) dx

3
...


8

 (x  2)(x 2  4) dx

dx

 sin x  sin 2x

2
...


 x
...
dx

 log x
...


3
...


1
 sin x
...


1 x 

cos 1 

 1  x 2 
...
x
...
dx

2

5
...




x
...
dx

LEVEL III
1
...


 1  cos 2x e

2  sin x

x

e x (1  x )

2
...



...


log x

 (1  log x) 2 dx


...
dx

(vi) Some Special Integrals
LEVEL I
1
...
dx

2
...
dx
LEVEL II

1
...
dx

2
...
dx
LEVEL III

1
...
dx

2
...


dx

 4 sin 2 x  5 cos 2 x (Hint: Divide the Numerator and Denominator by cos x and use the relation
2

sec2x=1+tan2x; and put tanx=t
LEVEL III
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SOLUTIONS: ASSIGNMENTS: INDEFINITE INTEGRALS
(i) Integration by substitution
LEVEL I
LEVEL II

1
...
sec x  C
3

1
...
2 tan x  C

1
3
...


3
...


tan 2 x
 log e tan x  C
2

2
...
 tan1cos x   C

1
sin 6x 
x  6   C
2


(ii) ) Application of trigonometric function in integrals
LEVEL I

1
...


3
1
cos x  cos 3x  C
4
12

x 1
1
1
 sin 6x  sin 4x  sin 2x  C
4 4
16
8

LEVEL II
LEVEL III

2
3

2
4
1
...
sin x  sin x  sin x  C

2
3

2
...


sin 3 x sin 5 x

C
3
5

(iii) Integration using Standard results

1
2

LEVEL I

1
...


LEVEL III

1
1
1 1 3x  2 
 x 1
4x 2  9  C 2
...
tan 
+ C
2
3
9
 3 
 3 

1
1
...
tan1sin x  2  C 3
...
x  log x  x  1 




2
2x  1
log
C
3
3

5
2

2
2
3
...
sin 1 x  1  x 2  C [Hint: Put x=cos2 ]

 2x  9 
2
  x  9x  20  C
2 


2
5
...
log x  1  2 log x  2  log x  3  C
2
2

LEVEL I

1
...
log
C

4
 x  3  2x  1

LEVEL II

1
...
log x  1 

3
8

1
4

2
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1 3
 log x  2  C
2x 4

LEVEL III





1
x
1
...

2
2

log 1  cos x  log 1  cos x  2 log1  2 cos x 


C
6
2
3





2
3
...
x
...
xlogx – x + C 3
...
logsecx + C





x 3 1
x2  2 1 x2
sin x 
C
3
9

1
...


3
...
2x tan1 x  log 1  x 2  C



5
...
tan x  logsec x  tan x   C
2

1
...

2x
2

x

LEVEL III

3
...
ex
...


=

e 2x
3 sin 3x  2 cos 3x   C
13

(vi) Some Special Integrals
LEVEL I

1
...


2
...

 sin 2x  C
2
2
4

x  2

x 2  4x  6
 log x  2  x 2  4x  6  C
2

x  2

1  4x  x 2 5 1 x  2 
 sin 
C
2
2
 5 

1
...




1
1 x  x2
3





3/ 2





1
2x  1 1  x  x 2  5 sin 1 2x  1   C


8
16
 5 

3 / 2 11
1 2
11
x x
 2x  1 x 2  x  log 2x  1  2 x 2  x   C



3
8
16 

(vii) Miscellaneous Questions

 2 tan x 
tan 1 
C


2 5
 5 
1

LEVEL II

1
...
+ log|cosx+sinx|+C

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(i) Integration by substitution

Text book , Vol
...
II Ex 7 Page 306,
Exercise 7
...
II Exp 8, 9, 10
Page 311,312,313, Exercise 7
...
II Exp 11&12
Page 318 Exp 13 319,Exp 14 & 15
Page320

(v) Integration by Parts

**

Text book , Vol
...
6 QNO ,10,11,
17,18,20

(vi)Some Special Integrals

Indefinit
e
Integrals

*

***

Text book , Vol
...
II Solved Ex
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