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 5:
Therefore, the anti derivative of
Question 2: Cos 3x
The anti derivative of cos 3x is a function of x
whose derivative is cos 3x.
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.
It is known that,
Ans: cot2x=cosec2x-1; antiderivative
of cot2 x is -cotx-x+c
Question 8: Find an anti-derivative
of √
with respect to x.
√
√
) = sinx+cosx
√(
Antiderivative of sinx+cosx is
cosx-sinx+c
Therefore, the anti derivative of
.
Question 9: Evaluate∫ (
Ans: e5x+c
Evaluate the following integrals:
Question 4:
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)
.
TWO MARK QUESTIONS: Evaluate the following integrals:
Write an antiderivative for each of the
following functions using
differentiation :
Question 1:
The anti derivative of
is the function
of x whose derivative is
It is known that,
Question 5:
.
Therefore, the anti derivative of
.
Question 2:
Question 6:
The anti derivative of
is the
function of x whose derivative is
.
It is known that,
Therefore, the anti derivative of
is
.
Evaluate the following integrals:
Question 7:
Question 3:
On dividing, we obtain
Question 4:
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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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Also,
3 mark questions:
If
such that f(2) = 0, find f(x)
Solution: It is given that
∴Anti derivative of
∴
INTEGRATION BY SUBSTITUTION:
ONE MARK QUESTIONS:
1`. Evaluate  tan 2 (2 x).dx
 tan
2

x
 dx .
2
2. Evaluate cos ec2 
(2 x).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.r.t x
Let 1 + log x = t
1.
Hint:
∴
=t
Ans: log(1+x2) +c
2.
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 9:
Let
∴ 2dx = dt
Question 5:
Let ax + b = t ⇒ adx = dt
Question 6:
Let 1 + 2x2 = t ∴ 4xdx = dt
Question 10:
Let
∴ 9x2 dx = dt
Question 7:
Let
∴ (2x + 1)dx = dt
Question 11:
Let log x = t ∴
Question 12:
Question 8:
Let
∴ −8x dx = dt
Let
∴
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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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Question 21:
Let 1 + cos x = t ∴ −sin x dx = dt
Question 24: Find
Let
∴
Question 25:
=
Question 22:
Let 1 + log x = t ∴
Question 23:
equals
Let
Also, let
⇒
THREE MARKS QUESTIONS
Integrate the following :
Question 2:
Let
Question 1:
Let
∴ 2adx = dt
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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 10:
Question 13:
Let
∴
Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt
Question 11:
∴
Let
Question 14:
Question 12:
Let
∴
Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt
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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 3: cos 2x cos 4x cos 6x
It is known that,
Question 6: sin x sin 2x sin 3x
It is known that
Question 4: sin3 (2x + 1)
Let
Question 7:sin 4x sin 8x
It is known that
Question 8:
Question 5: sin3 x cos3 x
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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 15:
Question 18:
Question 19:
Question 16: tan4x
From equation (1), we obtain
Question 20:
Question 17:
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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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INTEGRALS OF SOME PARTICULAR FUNCTIONS
TWO MARK QUESTIONS:
Question 5:
Integrate the following w.r.t x
Question 1:
Let x3 = t ∴ 3x2 dx = dt
Question 2:
Let 2x = t ∴ 2dx = dt
Question 6:
Let x3 = t ∴ 3x2 dx = dt
Question 3:
Let 2 − x = t ⇒ −dx = dt
Question 7:
Question 4:
Let 5x = t ∴ 5dx = dt
From (1), we obtain
Question 8:
Let x3 = t ⇒ 3x2 dx = dt
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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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THREE MAKRS QUESTIONS:
Question 3:
Question 1:
Question 4:
Question 2:
Equating the coefficients of x and
constant term on both sides, we obtain
4A = 4 ⇒ A = 1
A+B=1⇒B=0
Let 2x2 + x − 3 = t
∴ (4x + 1) dx = dt
Question 5:
Equating the coefficients of x and constant
term on both sides, we obtain
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From (1), we obtain
Question 6: Integrate
x2
x  2x  3
2
with respect to x.
I
1
 2x  2   1
2
dx  
dx
x 2  2x  3
x 2  2x  3
x2
1
2x  2
1
dx  
dx

2
2
2
x  2x  3
x  2x  3
1
1
  2 x 2  2x  3  
dx
2
2
x

1

2
 

 x 2  2x  3  log  x  1  x 2  2x  3  c
From equation (2), we obtain
ADDITIONAL 4 TO 5 MARK QUESTIONS: Integrate the following:
Question 1:
Equating the coefficient of x and constant term
on both sides, we obtain
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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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Question 5:
Equating the coefficients of x and constant
term on both sides, we obtain
Using equations (2) and (3) in (1), we obtain
Question 4:
Substituting (2) and (3) in (1), we obtain
Let x2 + 2x +3 = t
⇒ (2x + 2) dx =dt
Question 6:
Using equations (2) and (3) in (1), we obtain
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Equating the coefficients of x and constant
term, we obtain
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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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Equating the coefficients of x2, x, and constant
term, we obtain
A + C = 0 ;−A + B = 1 ; −B + C = 0
On solving these equations, we obtain
Question 3:
From equation (1), we obtain
Let
Substituting x = −1 and −2 in equation (1), we
obtain
A = −2 and B = 4
Question 4:
It can be seen that the given integrand is not a
proper fraction. 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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Let
Substituting x = 1 in equation (1), we obtain
B=4
Equating the coefficients of x2 and x, we
obtain A + C = 0; B − 2C = 3
Question 9:
It can be seen that the given integrand is not a
proper fraction.
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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Question 11:
[Hint: multiply numerator and denominator by
xn − 1 and put xn = t]
Question 13:
Let x2 = t ⇒ 2x dx = dt
Multiplying numerator and
denominator by xn − 1, we obtain
Substituting t = −3 and t = −1 in equation (1),
we obtain
Substituting t = 0, −1 in equation (1), we
obtain
A = 1 and B = −1
Question 14:
Multiplying numerator and denominator by x3,
we obtain
Question 12:
sin x = t]
[Hint: Put
Let x4 = t ⇒ 4x3dx = dt
Substituting t = 2 and then t = 1 in equation
(1), we obtain A = 1 and B = −1
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Substituting t = 0 and 1 in (1), we obtain
A = −1 and B = 1
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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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Question 2:
Equating the coefficient of x3, x2, x, and
constant term, we obtain
Equating the coefficients of x3, x2, x, and
constant term, we obtain
A + C = 0; B + D = 4 ;4A + 3C = 0
4B + 3D = 10
On solving these equations, we obtain
A = 0, B = −2, C = 0, and D = 6
On solving these equations, we obtain
Question 18:
INTEGRATION BY PARTS
TWO MARKS QUESTIONS:
Question 1: x sin x
Let I =
Taking x as first function and sin x as second
function and integrating by parts, we obtain
Question 3: ∫
Question 2:
.
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.r.t. x
Question 1:
Let
Taking x2 as first function and ex as second
function and integrating by parts, we obtain
Question 5: x log 2x
Again integrating by parts, we obtain
Let
Taking log 2x as first function and x as second
function and integrating by parts, we obtain
Question 2:
Let
Taking
as first function and x as
second function and integrating by parts, we
obtain
Question 6: x2 log x
Let
Taking log x as first function and x2 as second
function and integrating by parts, we obtain
Question 7:
Let
Taking x as first function and sec2x as second
function and integrating by parts, we obtain
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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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Question 7:
*******
Let
Taking
as first function and
as second function and integrating by parts, we
obtain
INTEGRAL OF THE FORM ∫
TWO MARK QUESTIONS
Question 1:
[ ( )
( )]
2. Evaluate:
e
x
 1  sin x 

 dx .
 1  cos x 
x
x

 1  2sin 2 cos 2 
I  e 
dx
2 x


2cos

2





1
x
  ex 
 tan  dx
x
2
 2cos 2

2

x
x
1
  e x  sec 2  tan  dx   e x  f '(x)  f (x)  dx
2
2
2
x
 e x tan
2
Let
x
Let
⇒
∴
It is known that,
Question 2:
THREE MARK QUESTIONS
Integrate the following w.r.t x
Also, let
It is known that,
⇒
Question 1:
Let
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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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Integrating by parts, we obtain
2. Find the integral of
with
√
respect to x and hence evaluate
∫
.
√
Solution: Let
√
Let
∴
then
∫√
∫
|
|
√
|
Question 5:
|
√
|
|
√
|
|
| |
,
| |
equals
Consider ∫
√
∫
Let
)
|(
⇒
Also, let
It is known that,
1. Find the integral of
)
√(
with respect
to x and evaluate ∫
.
∫
Consider
(
(
[
)(
)
(
)
(
)(
)
[
with
√
respect to x and hence evaluate
∫
.
√
Solution: Let
∴∫
√
then
Consider
[
∴
*(
∫
( )
∫
∫
Consider∫
(
√
+
|
|
√
(
)
[
∫√
∫
.
|
3. Find the integral of
Solution:
INDEFINITE INTEGRALS
(FIVE MARK QUESTIONS)
)
√(
)
]
(
]
)
(
]
)
*(
∫
|
)
|
+
|]
|
∫
(
Let
)
then
)
)
∴
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)
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|
|
.
4. Find the integral of
with respect
to x and hence evaluate ∫
Solution:
(
∫
Consider
(
(
[
)(
)
)
(
(
)(
)
[
∫
*(
)
[
|
|
+
∫
|
Let (
)
(
]
)
*(
)
=
(
|]
5x 2  2x
1
with
x  a2
and hence evaluate
2
1
dx ; Put x=atan ,
x  a2
2
then dx=a sec2 ;
|
(
(
|
∫
then
|
)
√
respect to
)
∫
respect to
1
|
√(
6.Find the integral of
+
|
.
|
x2+a2 =a2 tan2 +a2
)
|
)
5.Find the integral of

(
]
)
=∫
√
Let I= 
∫
=
=x2+2x+1=(x2+2x+1)+1
∫
)
]
|
Consider
∴
√
=(x+1)2+(1)2
[
∴
.
)
Now ∫
=a2(tan2 +1)=a2sec2
1
with
x a
and hence evaluate
2
dx
2
1
a sec 2 d
dx

 x2  a2
 a 2 sec2
1
1
x

 tan1  c
a
a
a
I

1
d
a
Consider ∫
Substituting
∫
√
=∫
√
=∫
√
(
∫√
)
7.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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√
applying
∫√
∫
√
∫√
.
√
|
√
√
|
|
√
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.
The students are advised to go through the
those questions also.
( )
(
=
+
Note: The above questions is for 5 mark
questions in part D of the question paper for
second PUC.
∫√
∫√
]
+ 2log| x+1+√
∫√
∫
∫
)
√
√
|
√
.
We get
√
Put x+1=t dx=dt
8. Find the integral of √
respect to x and evaluate
∫√
|
x2+2x+5=(x2+2x+1)+4=(x+1)2+4
√
|
√
|
√
|
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*******
Assignments
(i) Integration by substitution
LEVEL I
1
sec 2 (log x )
dx
1. 
x
e m tan x
2. 
dx
1 x2
3.

e sin
1 x
1 x2
dx
LEVEL II
1.

1
dx
x x
2.

1
x x6 1
dx
3.
1
 e x  1 dx
LEVEL III
1.
tan x
 sin x.cos x dx
2.
tan x
 sec x  cos x dx
3.
1
 sin x. cos 3 x dx
(ii) Application of trigonometric function in integrals
LEVEL I
1.
 sin
3
2
2.  cos 3x.dx
x.dx
3.
 cos x. cos 2x. cos 3x.dx
3.
 9x 2  12x  13
3.

3.

LEVEL II
1.
 sec
4
x. tan x.dx
2.

sin 4x
dx
sin x
LEVEL III
1.
 cos
5
x.dx
2.
 sin
2
x. cos 3 x.dx
(iii) Integration using standard results
LEVEL I
1.

dx
2.
4x 2  9
1
 x 2  2x  10 dx
dx
LEVEL II
1.
x
 x 4  x 2  1 dx
2.
cos x
 sin 2 x  4 sin x  5 dx
dx
7  6x  x 2
LEVEL III
1.

4.

2x
1 x  x
2
4
dx
1 x
dx
1 x
2.
x2  x 1
 x 2  x  1 dx
5.

6x  7
[CBSE 2011]
x  5x  4
(iv) Integration using Partial Fraction
LEVEL I
KHV’S
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x2
x  5x  6
2
dx
1.

2x  1
dx
( x  1)(x  1)

x  2x  8
dx
( x  1)( x  2)

2.
x2
dx
( x  1)( x  2)( x  3)
3.
3x  2
 (x  1) 2 (x  3) dx
LEVEL II
1.
x2  x 1
2
 x 2 (x  2)
2.
dx
x2 1
3.
 (x  1) 2 (x  3) dx
3.
 1  x 3 dx
LEVEL III
1.
8
 (x  2)(x 2  4) dx
dx
 sin x  sin 2x
2.
1
(v) Integration by Parts
LEVEL I
1.
 x.sec
e
x
2
x.dx
 log x.dx
2.
3.
(tan x  log sec x)dx
LEVEL II
1.
1
 sin x.dx

4.
1  1  x 


cos

 1  x 2 .dx


2
1
2. x . sin x.dx
2
5.
 sec
3
3.

x. sin 1 x
1 x
2
dx
x.dx
LEVEL III
1.
 coslog x dx
4.
 1  cos 2x e
2  sin x
x
.dx
e x (1  x )
2.
 (2  x) 2 dx
5.
e
2x
3.
log x
 (1  log x) 2 dx
. cos 3x.dx
(vi) Some Special Integrals
LEVEL I
1.

4  x 2 .dx
2.

1  4x 2 .dx
LEVEL II
1.

x 2  4x  6.dx
2.

1  4x  x 2 .dx
LEVEL III
1.
 (x  1)
1  x  x 2 .dx
2.
 (x  5)
x 2  x dx
(vii) Miscellaneous Questions
LEVEL II
1.
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
1.∫
KHV’S
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SOLUTIONS: ASSIGNMENTS: INDEFINITE INTEGRALS
(i) Integration by substitution
LEVEL I
1. tan(logex) + C
LEVEL II
1. 2 log e 1  x  C
LEVEL III
1. 2 tan x  C
1 m tan1 x
e
C
m
1 1 3
2. sec x  C
3
1
3. esin x  C
2.
3. log e 1  e x  C
2.  tan1cos x   C
3.
tan 2 x
 log e tan x  C
2
2.
1
sin 6x 
x
C

2
6 
(ii) ) Application of trigonometric function in integrals
LEVEL I
1. 
3.
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
2
4
1. 1 sec4 x  C OR tan x  tan x  C
4
2
4
LEVEL III
3
5
1. sin x  sin x  sin x  C
2
3
2
3
2. sin 3x  2 sin x  C
1
5
2.
sin 3 x sin 5 x

C
3
5
(iii) Integration using Standard results
1
2
LEVEL I
1. log e x 
LEVEL II
1.
LEVEL III
1
1. sin 
1
1
1 1 3x  2 
 x 1
4x 2  9  C 2. tan 1
 + C 3. tan 
+ C
2
3
9
 3 
 3 
 2x 2  1 
1
+C
tan 1


3
3


 2x  1 
C
 5 
1
2. tan1sin x  2  C 3. sin 
 2x 2  1 
C

 5 
1
2
2
2. x  log x  x  1 


2
2x  1
log
C
3
3
5
2
2
2
3. x  5x  6  log  x    x  5x  6  C
4. sin 1 x  1  x 2  C [Hint: Put x=cos2 ]
 2x  9 
2
  x  9x  20  C
2


2
5. 6 x  9x  20  34 log 
(iv) Integration using Partial Fraction
1
3
2. log x  1  2 log x  2  log x  3  C
2
2
LEVEL I
1. 1 log( x  1)  5 log( x  2)  C
3
3
11  x  1 
5
3. log
C

4
 x  3  2x  1
LEVEL II
1. x – 11log(x – 1) + 16log(x – 2) + C
3. log x  1 
3
8
1
4
2. log x 
1
5
 log x  3  C
2x  1 8
KHV’S
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1 3
 log x  2  C
2x 4
LEVEL III


1
x
1. log(x +2)  log x 2  4  tan 1
2.
2
2
log 1  cos x  log 1  cos x  2 log1  2 cos x 


C
6
2
3


2
3. log 1  x   log 1  x  x 
1
3
1
6
(v) Integration by Parts
LEVEL I
1.x.tanx + logcosx + C
LEVEL II
1
 2x  1 
tan 1
 + C [Hint: Partial fractions]
3
 3 
2.xlogx – x + C 3.ex.logsecx + C


x 3 1
x2  2 1 x2
sin x 
C
3
9
1. x sin 1 x  1  x 2  C
2.
3.  1  x 2 sin 1 x  x  C
4. 2x tan1 x  log 1  x 2  C

5.
1
sec x. tan x  logsec x  tan x   C
2
1.
e
x
 C [Hint:
coslog x   sinlog x   C 2.
2x
2

x
LEVEL III
3.
x
C
1  log x
4. ex.tanx + C
5.
=
e 2x
3 sin 3x  2 cos 3x   C
13
(vi) Some Special Integrals
LEVEL I
1.
LEVEL II
1.
2.
LEVEL III
x 4  x2
x 1  4x 2 1 1
 2 log x  4  x 2  C 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. 
2.

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.
LEVEL III
1. + log|cosx+sinx|+C
KHV’S
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Detail of the concepts to be mastered by every student of class second PUC
with exercises and examples of NCERT Text Book.
Indefinit
e
Integrals
(i) Integration by substitution
*
Text book , Vol. II Examples 5&6
Page 300, 302,301,303
(ii) ) Application of trigonometric
function in integrals
**
Text book , Vol. II Ex 7 Page 306,
Exercise 7.3 Q13&Q24
(iii) Integration of some particular
function
***
Text book , Vol. II Exp 8, 9, 10
Page 311,312,313, Exercise 7.4 Q
3,4,8,9,13&23
(iv) Integration using Partial Fraction
***
Text book , Vol. II Exp 11&12
Page 318 Exp 13 319,Exp 14 & 15
Page320
(v) Integration by Parts
**
Text book , Vol. II Exp 18,19&20
Page 325 Exs 7.6 QNO ,10,11,
17,18,20
(vi)Some Special Integrals
***
Text book , Vol. II Exp 23 &24
Page 329
**
Text book , Vol. II Solved Ex. 40,
41
dx
 x2  a2 
dx




,
1
a2  x2
x a
2
2
,
dx
 ax 2  bx  c ,
dx ,
dx
ax 2  bx  c
(px  q)dx
,
(px  q)dx
 ax 2  bx  c ,
ax 2  bx  c
a 2  x 2 dx ,

x 2  a 2 dx
(vii) Miscellaneous Questions
viii)Some special integrals
Text book Supplimentary material
Page 614,615
SYMBOLS USED :
* : Important Questions, ** :Very Important Questions, *** : Very-Very Important Questions
KHV’S
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