Expertbulletin.com provides you step by step NCERT Solutions for Class 12 Maths Chapter 7 Integrals solved by Experts as per NCERT (CBSE) Book guidelines. All Relations and Functions Exercise Questions with Solutions to help you to score high marks in upcoming exams and other competitive exams. following topics and sub-topics are covered in the Chapter 7 Integrals
7.1 Introduction
7.2 Integration as an Inverse Process of Differentiation
7.2.1 Geometrical interpretation of indefinite integral
7.2.2 Some properties of indefinite integral
7.2.3 Comparison between differentiation and integration
7.3 Methods of Integration
7.3.1 Integration by substitution
7.3.2 Integration using trigonometric identities
7.4 Integrals of Some Particular Functions
7.5 Integration by Partial Fractions
7.6 Integration by Parts
7.6.1 Integral of the type
7.6.2 Integrals of some more types
7.7 Definite Integral
7.7.1 Definite integral as the limit of a sum
7.8 Fundamental Theorem of Calculus
7.8.1 Area function
7.8.2 First fundamental theorem of integral calculus
7.8.3 Second fundamental theorem of integral calculus
7.9 Evaluation of Definite Integrals by Substitution
7.10 Some Properties of Definite Integrals.

## Integrals NCERT Solutions – Class 12 Maths

### NCERT Solutions for Class 12 Maths Chapter 7

Exercise 7.1 : Solutions of Questions on Page Number : 299
Q1 :sin 2x
Answer :The anti derivative of sin 2x is a function of x whose derivative is sin 2x.
It is known that,

Therefore, the anti derivative of

Q2 :Cos 3x
Answer : The anti derivative of cos 3x is a function of x whose derivative is cos 3x.
It is known that,

Therefore, the anti derivative of .

Q3 :e2x
Answer :The anti derivative of e2x is the function of x whose derivative is e2x.
It is known that,

Therefore, the anti derivative of .

Q4 :Answer :The anti derivative of is the function of x whose derivative is .
It is known that,

Therefore, the anti derivative of

Q5 :
The anti derivative of is the function of x whose derivative is
It is known that,

Therefore, the anti derivative of is .

Q7 :

Q8 :

Q9 :

Q10 :

Q11 :

Q12 :

Q13 :

On dividing, we obtain

Q15 :

Q16 :

Q21 :The anti derivative of equals
(A) (B)
(C) (D)

Hence, the correct answer is C.

Q22 :If such that f(2) = 0, then f(x) is
(A) (B)
(C) (D)
Answer :It is given that, ∴Anti derivative of
Also,
Hence, the correct answer is A.

Exercise 7.2 : Solutions of Questions on Page Number : 304

Q1 :
Let = t
∴2x dx = dt

Let log |x| = t

Q3 :
Let 1 + log x = t

Q4 :sin x . sin (cos x)
Answer : sin x .sin (cos x)
Let cos x = t
∴ – sin x dx = dt

Let ax + b = t

Q7 :Answer :Let ∴ dx = dt

Let 1 + 2x2 = t
∴ 4xdx = dt

Q9 :
Let
∴ (2x + 1)dx = dt

Q11 :

Q12 :
Let

Q13 :
Let∴ 9×2 dx = dt

Let log x = t

Let∴ – 8x dx = dt

Let∴ 2dx = dt

Let∴ 2xdx = dt

Let

Q19 :
Dividing numerator and denominator by ex, we obtain
Let

Let

Q21 :

Let 2x – 3 = t
∴ 2dx = dt

Let 7 – 4x = t
∴ – 4dx = dt

Let

Q24 :

Q26 :
Let

Q27 :
Let sin 2x = t

Q28 :
Let∴ cos x dx = dt

Q29 :cot x log sin x
Let log sin x = t

Q30 :
Let 1 + cos x = t
∴ – sin x dx = dt

Q31 :
Let 1 + cos x = t
∴ – sin x dx = dt

Q32 :

Let sin x + cos x = t ⇒ (cos x – sin x) dx = dt

Q33 :

Put cos x – sin x = t ⇒ ( – sin x – cos x) dx = dt

Q34 :

Q35 :
Let 1 + log x = t

Q36 :
Let

Q37 :
Let x4 = t
∴ 4x3dx = dt

Let

From (1), we obtain

Q38 :equals

Let

Hence, the correct answer is D.

Q39 :equals

Let

Hence, the correct answer is D.

Exercise 7.3 : Solutions of Questions on Page Number : 307

Q1 :

Q2 :
It is known that,

Q3 : cos 2x cos 4x cos 6x
It is known that,

Q4 : sin3 (2x + 1)
Let

Q5 : sin3 x cos3 x

Q6 : sin x sin 2x sin 3x
It is known that,

Q7 : sin 4x sin 8x
It is known that,

Q8 :

Q9 :

Q10 : sin4 x

Q11 : cos4 2x

Q12 :

Q13 :

Q14 :

Q15 :

Q16 : tan4x

From equation (1), we obtain

Q17 :

Q18 :

Q19 :

Q20 :

Q21 : sin-1 (cos x)

It is known that,

Substituting in equation (1), we obtain

Q22 :

Q23 :is equal to
A. tan x + cot x + C
B. tan x + cosec x + C
C. – tan x + cot x + C
D. tan x + sec x + C

Hence, the correct answer is A.

Q24 :equals
A. – cot (exx) + C
B. tan (xex) + C
C. tan (ex) + C
D. cot (ex) + C
Let exx = t

Hence, the correct answer is B.

Exercise 7.4 : Solutions of Questions on Page Number : 315

Q1 :
Let x3 = t
∴ 3x2 dx = dt

Q2 :
Let 2x = t
∴ 2dx = dt

Q3 :
Let 2 – x = t
⇒ – dx = dt

Q4 :
Let 5x = t
∴ 5dx = dt

Q5 :

Q6 :
Let x3 = t
∴ 3×2 dx = dt

Q7 :

From (1), we obtain

Q8 :
Let x3 = t
⇒ 3x2 dx = dt

Q9 :
Let tan x = t
∴ sec2x dx = dt

Q10 :

Q12 :

Q13 :

Q14 :

Q15 :

Q16 :

Equating the coefficients of x and constant term on both sides, we obtain
4A = 4 ⇒ A = 1
A + B = 1 ⇒ B = 0
Let 2×2 + x – 3 = t
∴ (4x + 1) dx = dt

Q17 :

Equating the coefficients of x and constant term on both sides, we obtain

From (1), we obtain

From equation (2), we obtain

Q18 :

Equating the coefficients of x and constant term on both sides, we obtain

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

Q19 :

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

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

Q20 :

Equating the coefficients of x and constant term on both sides, we obtain

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

Q21 :

Let x2 + 2x +3 = t
⇒ (2x + 2) dx =dt

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

Q22 :

Equating the coefficients of x and constant term on both sides, we obtain

Substituting (2) and (3) in (1), we obtain

Q23 :

Equating the coefficients of x and constant term, we obtain

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

Q24 :equals
A. x tan – 1 (x + 1) + C
B. tan – 1 (x + 1) + C
C. (x + 1) tan – 1 x + C
D. tan – 1x + C

Hence, the correct answer is B.

Q25 :equals
A.
B.
C.
D.

Hence, the correct answer is B.

Exercise 7.5 : Solutions of Questions on Page Number : 322

Q1 :
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

Q2 :
Let

Equating the coefficients of x and constant term, we obtain
A + B = 0
– 3A + 3B = 1
On solving, we obtain

Q3 :
Let

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

Q4 :
Let
Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

Q5 :
Let
Substituting x = – 1 and – 2 in equation (1),we obtain
A = – 2 and B = 4

Q6 :
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

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

Substituting in equation (1), we obtain

Q7 :
Let

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

From equation (1), we obtain

Q8 :
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

Q9 :
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
On solving, we obtain

Q10 :

Let

Equating the coefficients of x2 and x, we obtain

Q11 :

Let

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

Q12 :
It can be seen that the given integrand is not a proper fraction.
Therefore, on dividing (x3 + x + 1) by x2 – 1, we obtain

Let

Substituting x = 1 and – 1 in equation (1), we obtain

Q13 :

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

Q14 :

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

Q15 :

Equating the coefficient of x3, x2, x, and constant term, we obtain

On solving these equations, we obtain

Q16 :[Hint: multiply numerator and denominator by xn – 1 and put xn = t]

Multiplying numerator and denominator by xn – 1, we obtain

Substituting t = 0, – 1 in equation (1), we obtain
A = 1 and B = – 1

Q17 :[Hint: Put sin x = t]

Substituting t = 2 and then t = 1 in equation (1), we obtain
A = 1 and B = – 1

Q18 :

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

Q19 :

Let x2 = t ⇒ 2x dx = dt

Substituting t = – 3 and t = – 1 in equation (1), we obtain

Q20 :

Multiplying numerator and denominator by x3, we obtain

Let x4 = t ⇒ 4x3dx = dt

Substituting t = 0 and 1 in (1), we obtain
A = – 1 and B = 1

Q21 :[Hint: Put ex = t]

Let ex = t ⇒ ex dx = dt

Substituting t = 1 and t = 0 in equation (1), we obtain
A = – 1 and B = 1

Q22 :
A.
B.
C.
D.

Substituting x = 1 and 2 in (1), we obtain
A = – 1 and B = 2

Hence, the correct answer is B.

Q23 :
A.
B.
C.
D.

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

Hence, the correct answer is A.

Exercise 7.6 : Solutions of Questions on Page Number : 327

Q1 : x sin x
Let I =
Taking x as first function and sin x as second function and integrating by parts, we obtain

Q2 :
Let I =
Taking x as first function and sin 3x as second function and integrating by parts, we obtain

Q3 :
Let
Taking x2 as first function and ex as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

Q4 : x logx
Let
Taking log x as first function and x as second function and integrating by parts, we obtain

Q5 :x log 2x
LetTaking log 2x as first function and x as second function and integrating by parts, we obtain

Q6 : x2 log x
LetTaking log x as first function and x2 as second function and integrating by parts, we obtain

Q7 :
Let
Taking as first function and x as second function and integrating by parts, we obtain

Q8 :
Let
Taking as first function and x as second function and integrating by parts, we obtain

Q9 :
Let
Taking cos – 1 x as first function and x as second function and integrating by parts, we obtain

Q10 :
Let
Taking as first function and 1 as second function and integrating by parts, we obtain

Q11 :
Let

Taking as first function and as second function and integrating by parts, we obtain

Q12 :
Let
Taking x as first function and sec2x as second function and integrating by parts, we obtain

Q13 :
Let
Taking as first function and 1 as second function and integrating by parts, we obtain

Q14 :
Taking as first function and x as second function and integrating by parts, we obtain

Q15 :
Let
Let I = I1 + I2 … (1)
Where,and
Taking log x as first function and x2 as second function and integrating by parts, we obtain

Taking log x as first function and 1 as second function and integrating by parts, we obtain

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

Q16 :
Let
Let

It is known that,

Q17 :
Let

Let

It is known that,

Q18 :

Let
It is known that,
From equation (1), we obtain

Q19 :
Also, let
It is known that,

Q20 :

Let

It is known that,

Q21:
Let
Integrating by parts, we obtain

Again integrating by parts, we obtain

Q22 :
Let
= 2θ

Integrating by parts, we obtain

Q23 :equals

Let
Also, let

Hence, the correct answer is A.

Q24 :equals

Let
Also, let
It is known that,

Hence, the correct answer is B.

Exercise 7.7 : Solutions of Questions on Page Number : 330

Q1 :

Q2 :

Q3 :

Q4 :

Q5 :

Q6 :

Q7 :

Q8 :

Q9 :

Q10 :is equal to
A.
B.
C.
D.

Hence, the correct answer is A.

Q11 :is equal to
A.
B.
C.
D.

Hence, the correct answer is D.

Exercise 7.8 : Solutions of Questions on Page Number : 334

Q1 :
It is known that,

Q2 :
It is known that,

Q3 :
It is known that,

Q4 :

It is known that,

From equations (2) and (3), we obtain

Q5 :

It is known that,

Q6 :
It is known that,

Exercise 7.9 : Solutions of Questions on Page Number : 338

Q1 :

By second fundamental theorem of calculus, we obtain

Q2 :

By second fundamental theorem of calculus, we obtain

Q3 :

By second fundamental theorem of calculus, we obtain

Q4 :

By second fundamental theorem of calculus, we obtain

Q5 :

By second fundamental theorem of calculus, we obtain

Q6 :

By second fundamental theorem of calculus, we obtain

Q7 :

By second fundamental theorem of calculus, we obtain

Q8 :

By second fundamental theorem of calculus, we obtain

Q9 :

By second fundamental theorem of calculus, we obtain

Q10 :

By second fundamental theorem of calculus, we obtain

Q11 :

By second fundamental theorem of calculus, we obtain

Q12 :

By second fundamental theorem of calculus, we obtain

Q13 :

By second fundamental theorem of calculus, we obtain

Q14 :

By second fundamental theorem of calculus, we obtain

Q15 :

By second fundamental theorem of calculus, we obtain

Q16 :
Let
Equating the coefficients of x and constant term, we obtain
A = 10 and B = – 25

Substituting the value of I1 in (1), we obtain

Q17 :

By second fundamental theorem of calculus, we obtain

Q18 :

By second fundamental theorem of calculus, we obtain

Q19 :

By second fundamental theorem of calculus, we obtain

Q20 :

By second fundamental theorem of calculus, we obtain

Q21 : equals
A.
B.
C.
D.

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

Q22 : equals
A.
B.
C.
D.

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is C.

Exercise 7.10 : Solutions of Questions on Page Number : 340

Q1 :

When x = 0, t = 1 and when x = 1, t = 2

Q2 :

When x = 0, t = 1 and when x = 1, t = 2

Q3 :

Also, let

Q4 :

Also, let

Q5 :

Also, let x = tanθ ⇒ dx = sec2θ dθ
When x = 0, θ = 0 and when x = 1,

Takingθas first function and sec2θ as second function and integrating by parts, we obtain

Q6 :

Also, let x = tanθ ⇒ dx = sec2θ dθ
When x = 0, θ = 0 and when x = 1,
Taking θ as first function and sec2θ as second function and integrating by parts, we obtain

Q7 :

Let x + 2 = t2 ⇒ dx = 2tdt
When x = 0, and when x = 2, t = 2

Q8 :

Let x + 2 = t2 ⇒ dx = 2tdt
When x = 0, and when x = 2, t = 2

Q9 :

Let cos x = t ⇒ – sinx dx = dt
When x = 0, t = 1 and when

Q10 :

Let cos x = t ⇒ – sinx dx = dt
When x = 0, t = 1 and when

Q11 :

Let ⇒ dx = dt

Q12 :

Let ⇒ dx = dt

Q13 :

Let x + 1 = t ⇒ dx = dt
When x = – 1, t = 0 and when x = 1, t = 2

Q14 :

Let x + 1 = t ⇒ dx = dt
When x = – 1, t = 0 and when x = 1, t = 2

Q15 :

Let 2x = t ⇒ 2dx = dt
When x = 1, t = 2 and when x = 2, t = 4

Q16 :
Let 2x = t ⇒ 2dx = dt
When x = 1, t = 2 and when x = 2, t = 4

Q17 : The value of the integral is
A. 6
B. 0
C. 3
D. 4

Let cotθ = t ⇒ – cosec2θ dθ= dt

Hence, the correct answer is A.

Q18 : The value of the integral is
A. 6
B. 0
C. 3
D. 4

Let cotθ = t ⇒ – cosec2θ dθ= dt

Hence, the correct answer is A.

Q19 : If
A. cos x + x sin x
B. x sinx
C. x cos x
D. sin x + x cos x

Integrating by parts, we obtain

Hence, the correct answer is B.

Q20 : If
A. cos x + x sin x
B. x sinx
C. x cos x
D. sin x + x cos x

Integrating by parts, we obtain

Hence, the correct answer is B.

Exercise 7.11 : Solutions of Questions on Page Number : 347

Q1 :

Adding (1) and (2), we obtain

Q2 :

Adding (1) and (2), we obtain

Q3 :

Adding (1) and (2), we obtain

Q4 :

Adding (1) and (2), we obtain

Adding (1) and (2), we obtain

Q6 :
Adding (1) and (2), we obtain

Adding (1) and (2), we obtain

Q9 :

Q10 :

Adding (1) and (2), We obtain

As sin2(-x)=(sin(-x)) = (-sinx)2= sin2x, therefore, sin2x is an even function.
It is known that if f(x) is an even function,then

Q12 :

Q13 :

As sin7(-x)=(sin(-x))7=(-sinx)7=-sin7x,therefore,sin2x is an even function.
It is known that if f(x) is an even function,then

Q14 :
It is known that,

Q15 :

Adding (1) and (2), we obtain,

Q16 :

Adding (1) and (2), we obtain,
sin(Π-x)=sinx

Adding (4) and (5), we obtain,

Q17 :
It is known that,

Q18 :
It can be seen that,(x-a) ≤ 0 when 0 ≤ x ≤1 and(x-1) ≥0 when 1 ≤ x ≤ 4.

Q19 : Show that if f and g defined asand

Adding (1) and (2), we obtain

Q20 : The value of
A
.0
B.2
C.Π
D.

It is known that if f(x) is an even function,then
and f(x) is an odd function,then

Q21 : The value of
A. 2
B. C.
0
D.
-2

Hence the correct answer is C.

Exercise Miscellaneous : Solutions of Questions on Page Number : 352

Q1 :

Equating the coefficients of x2, x, and constant term, we obtain
– A + B – C = 0
B + C = 0
A = 1
On solving these equations, we obtain
From equation (1), we obtain

Q2 :

Q3 :
[Hint: Put]

Q4 :

Q5 :

On dividing, we obtain

Q6 :

Equating the coefficients of x2, x, and constant term, we obtain
A + B = 0
B + C = 5
9A + C = 0
On solving these equations, we obtain
From equation (1), we obtain

Q7 :
Let x – a = t ⇒ dx = dt

Q8 :

Q9 :
Let sin x = t ⇒ cos x dx = dt

Q10 :

Q11 :

Q12 :
Let x4 =t ⇒ 4x3 dx = dt

Q13 :
Let ex = t ⇒ ex dx = dt

Q14 :

Equating the coefficients of x3, x2, x, and constant term, we obtain
A + C = 0
B + D = 0
4A + C = 0
4B + D = 1
On solving these equations, we obtain

From equation (1), we obtain

Q15 :

= cos3 x × sin x
Let cos x = t ⇒ – sin x dx =dt

Q16 :

Q17 :

Q18 :

Q19 :

Q20 :

Q21 :

Q22 :

Equating the coefficients of x2, x,and constant term, we obtain
A + C = 1
3A + B + 2C = 1
2A + 2B + C = 1
On solving these equations, we obtain
A = – 2, B = 1, and C = 3
From equation (1), we obtain

Q23 :

Q24 :

Integrating by parts, we obtain

Q25 :

Q26 :

When x = 0, t = 0 and

Q27 :
When and when

Q28 :

When and when
As ,
therefore, is an even function.
It is known that if f(x) is an even function, then

Q29 :

Q30 :

Q31 :

From equation (1), we obtain

Q32 :

Adding (1) and (2), we obtain

Q33 :

a
From equation (1),(2),(3) and (4), we obtain

Q34 :

Equating the coefficients of x2, x, and constant term, we obtain
A + C = 0
A + B = 0
B = 1
On solving these equations, we obtain
A = – 1, C = 1, and B = 1
Hence, the given result is proved.

Integrating by parts, we obtain

Hence, the given result is proved.

Q36 :

Therefore, f (x) is an odd function.
It is known that if f(x) is an odd function, then

Hence, the given result is proved.

Q37 :

Hence, the given result is proved.

Hence, the given result is proved.

Q39 :
Integrating by parts, we obtain

Let 1 – x2 = t ⇒ – 2x dx = dt

Hence, the given result is proved.

Q40 :
Evaluate as a limit of a sum.
It is known that,

Q41 :
is equal to
A.
B.
C.
D.

Hence, the correct answer is A.

Q42 :
is equal to

A.
B.
C.
D.

Hence, the correct answer is B.

Q43 : If then is equal to
A.
B.
C.
D.
Hence, the cor

Q44 :
The value of is
A. 1
B. 0
C. – 1
D.