# NCERT Solutions For Class 12 Maths Chapter 7 – Integrals Class 12 NCERT Solutions

NCERT Solutions For Class 12 Maths Chapter 7 clears all the doubts regarding the chapter on Integrals and Questions are solved in a simple way. Students who are appearing for CBSE Class 12 board exams can download Chapter 7 Integrals Class 12 NCERT Solutions. In this chapter, each question was solved by the best teachers in India. All answers are given with a complete explanation. This will helps to get a better score in the examination. Class 12 Maths NCERT Solutions Chapter 7 will give you the best results in the CBSE board exams. Maths is the most important and scoring subject in the board exams. There are 10 exercises in NCERT Solutions For Class 12 Maths Chapter 7. Experts are given detailed explanations for the students. NCERT Solutions For Class 12 is useful for the students who are going to attempt JEE Mains, JEE Advance. It will help to get the best score in the exam and you can complete your home assignments as well as revision purposes.

## NCERT Solutions Class 12 Maths Chapter 7 Integrals

 Section no Topic Name 7.1 Introduction 7.2 Integration as an Inverse Process of Differentiation 7.3 Methods of Integration 7.4 Integrals of Some Particular Functions 7.5 Integration by Partial Fractions 7.6 Integration by Parts 7.7 Definite Integral 7.8 Fundamental Theorem of Calculus 7.9 Evaluation of Definite Integrals by Substitution 7.10 Some Properties of Definite Integrals

### NCERT Solutions For Class 12 Maths Chapter 7 Ex 7.1 Introduction

Question 1:
Sin 2x

The antiderivative of sin 2x is a function of x whose derivative is sin 2x.

It is known that,

Therefore, the antiderivative of sin 2x is

Question 2:
Cos 3x
The antiderivative of cos 3x is a function of x whose derivative is cos 3x
It is known that,

Therefore, the antiderivative of cos 3x is

Question 3:
e2x

The antiderivative of e2is the function of x whose derivative is e2x.

It is known that,

Therefore, the antiderivative of

Question 4:

The antiderivative of
is the function of whose derivative is
It is known that,

Therefore, the antiderivative of is

Question 5:

The antiderivative of is the function of x whose derivative is
It is known that,

Therefore, the antiderivative of is

Question 6:

Question 7:

Question 8:

Question 9:

Question 10:

Question 11:

Question 12:

Question 13:

On dividing, we obtain

Question 14:

Question 15:

Question 16:

Question 17:

Question 18:

Question 19:

Question 20:

Question 21:
The antiderivative of equals
(A)(B) (C)(D)

Hence, the correct answer is C.

Question 22:
If such that f(2) = 0, then f(x) is
(A)(B)(C)(D)
It is given that

∴Antiderivative of

also
f(2)=0

Hence, the correct answer is A.

### NCERT Solutions For Class 12 Maths Chapter 7 Integrals Ex 7.2 Integration as an Inverse Process of Differentiation

Question 1:

Let =t
∴2x dx = dt

Question 2:

Let log |x| = t

Question 3:

Let 1 + log t

Question 4:
Sin x.Sin (Cos x)

in x ⋅ sin (cos x)

Let cos x = t
∴ −sin x dx = dt

Question 5:
sin(ax+b)cos(ax+b)

let 2(ax+b)=t

Question 6:

Let ax + b = t

Question 7:

let (x+2)=t
∴ dx = dt

Question 8:

Let 1 + 2x2 = t

∴ 4xdx = dt

Question 9:

Let
∴ (2x + 1)dx = dt

Question 10:

let

Question 11:

Let x+4=t
⇒dx=dt,x=t-4

Question 12:

Let

Question 13:

Let
∴ 9x2 dx = dt

Question 14:

Let log x = t

Question 15:

Let
∴ −8x dx = dt

Question 16:

Let 2x+3=t
∴ 2dx = dt

Question 17:

Let
∴ 2xdx = dt

Question 18:

Let

Question 19:

Dividing the numerator and denominator by ex, we obtain

Let

Question 20:

Let

Question 21:

Let 2x − 3 = t

∴ 2dx = dt

⇒∫tan22x-3dx = ∫sec22x-3 – 1dx=∫sec2t- 1dt2= 12∫sec2t dt – ∫1dt= 12tant – t + C= 12tan2x-3 – 2x-3 + C

Question 22:

Let 7 − 4x = t

∴ −4dx = dt

Question 23:

Let

Question 24:

Let 3 cos x+ 2 sin x =t
∴(-3 sin x +2 cos x)dx=dt

Question 25:

Let (1-tan x)=t

Question 26:

Let

Question 27:

Let sin 2x = t
2 cos 2x dx=dt

Question 28:

Let 1+sin x=t
∴ cos x dx = dt

Question 29:
cot x log sin x
Let log sin x = t

Question 30:

Let 1 + cos x = t

∴ −sin x dx = dt

Question 31:

Let 1 + cos x = t

∴ −sin x dx = dt

Question 32:

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

Question 33:

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

Question 34:

Question 35:

Let 1 + log x = t

Question 36:

Let (x+log x)=t

Question 37:

Let x4 = t

∴ 4x3 dx = dt

Let

From (1), we obtain

Question 38:
equals
(A)(B)(C)(D)
Let

Question 39:
equals
A. tan x +cot x+C
B. tan x-cot x+C
C.tan x cot x+C
D. tan x-cot 2x+C

Hence, the correct answer is B.

### NCERT Solutions For Class 12 Matas Chapter 7 Ex 7.3 Methods of Integration

Question 1:

Question 2:
sin 3x cos 4x
It is known that,

Question 3:
cos 2x cos 4x cos 6x
It is known that,

Question 4:

Let

Question 5:

Question 6:
sin x sin 2x sin 3x
It is known that,

Question 7:
sin 4x sin 8x

It is known that,

sin A . sin B = 12cosA-B-cosA+B∴∫sin4x sin8x dx=∫12cos4x-8x-cos4x+8xdx=12∫cos-4x-cos12xdx=12∫cos4x-cos12xdx=12sin4x4-sin12x12+C

Question 8:

Question 9:

Question 10:
sin4x

Question 11:
cos4 2x

Question 12:

Question 13:

Question 14:

Question 15:

Question 16:

From equation (1), we obtain

Question 17:

Question 18:

Question 19:

1sinxcos3x=sin2x+cos2xsinxcos3x=sinxcos3x+1sinxcosx

⇒1sinxcos3x=tanxsec2x+1cos2xsinxcosxcos2x=tanxsec2x+sec2xtanx

Question 20:

Question 21:
sin−1 (cos x)

It is known that,

Substituting in equation (1), we obtain

Question 22:

Question 23:
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.

Question 24:
equals

A. âˆ’ cot (exx) + C

B. tan (xex) + C

C. tan (ex) + C

D. cot (ex) + C

Let exx = t

### NCERT Solutions For Class 12 Matas Chapter 7 Ex 7.4 Integrals of Some Particular Functions

Question 1:

Let x3 = t

∴ 3x2 dx = dt

Question 2:

Let 2x = t

∴ 2dx = dt

Question 3:

Let 2 − t

⇒ −dx = dt

Question 4:

Let 5x = t

∴ 5dx = dt

Question 5:

Question 6:

Let x3 = t

∴ 3x2 dx = dt

Question 7:

From (1), we obtain

Question 8:

Let x3 = t

⇒ 3x2 dx = dt

Question 9:

Let tan x = t

∴ sec2x dx = dt

Question 10:

Question 11:
19×2+6x+5

∫19×2+6x+5dx=∫13x+12+22dx
Let (3x+1)=t
∴3 dx=dt
⇒∫13x+12+22dx=13∫1t2+22dt
=13×2tan-1t2+C
=16tan-13x+12+C

Question 12:

Question 13:

Question 14:

Question 15:

Question 16:

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 17:

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

From (1), we obtain

From equation (2), we obtain

Question 18:

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

therefore

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

Question 19:

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

Question 20:

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

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

Question 21:

Let x2 + 2x +3 = t

⇒ (2x + 2) dx =dt

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

Question 22:

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

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

Question 23:

Equating the coefficients of x and constant term, we obtain

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

Question 24:

equals

A. x tan−1 (x + 1) + C

B. tan− 1 (x + 1) + C

C. (x + 1) tan−1 x + C

D. tan−1 x + C

Hence, the correct answer is B.

Question 25:
equals
A.
B.
C.
D.

Hence, the correct answer is B.

### NCERT Solutions Class 12 Maths Chapter 7 Integrals Ex 7.5 Integration by Partial Fractions

Question 1:

Let
⇒x=A(x+2)+B(x+1)

Equating the coefficients of x and constant term, we obtain

A + = 1

2A + B = 0

On solving, we obtain

A = −1 and B = 2

Question 2:

Let
1=A(x-3)+B(x+3)

Equating the coefficients of x and constant term, we obtain

A + B = 0

−3A + 3B = 1

On solving, we obtain

Question 3:

Let
3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)……..(1)

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

Question 4:

Let
x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)………(1)
Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

Question 5:

Let
2x=A(x+2)+B(x+1)…….(1)

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

A = −2 and B = 4

Question 6:

It can be seen that the given integrand is not a proper fraction.

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

Substituting x = 0 and

in equation (1), we obtain

= 2 and B = 3

Substituting equation (1), we obtain

Question 7:

Let

Equating the coefficients of x2x, 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

Question 8:

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 9:

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

Question 10:

Let

Question 11:

Let

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

Question 12:

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
2x+1=A(x-1)+B(x+1)……..(1)
Substituting = 1 and −1 in equation (1), we obtain

Question 13:

Equating the coefficient of x2x, 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

Question 14:

Equating the coefficient of x and constant term, we obtain

A = 3

2A + = −1 ⇒ B = −7

Question 15:

Equating the coefficient of x3x2, x, and constant term, we obtain

By solving these equations, we obtain

Question 16:
[Hint: multiply numerator and denominator by xn − 1 and put xn = t] Answer:

Multiplying the numerator and denominator by x− 1, we obtain

Substituting t = 0, −1 in equation (1), we obtain

A = 1 and B = −1

Question 17:
[Hint: Put sin x = t] Answer:

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

A = 1 and B = −1

Question 18:

Equating the coefficients of x3x2x, 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

Question 19:

Let x2 = t ⇒ 2xdx = dt

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

Question 20:

Multiplying the 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

Question 21:
[Hint: Put ex = t] Answer:

Let ex = ⇒ exdx = dt

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

A = −1 and B = 1

Question 22:

A.
B.
C.
D.

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

A = −1 and B = 2

Hence, the correct answer is B.

Question 23:

A.
B.
C.
D.

Equating the coefficients of x2x, and constant term, we obtain

A + B = 0

C = 0

A = 1

On solving these equations, we obtain

= 1, B = −1, and C = 0

Hence, the correct answer is A.

### NCERT Solutions Class 12 Maths Chapter 7 Integrals Ex 7.6 Integration by Parts

Question 1:
x sin x
Let l=
Taking x as first function and sin x as second function and integrating by parts, we obtain

Question 2:
x sin 3x
Let l=
Taking x as first function and sin 3x as second function and integrating by parts, we obtain

Question 3:

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

Again integrating by parts, we obtain

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

Question 5:
x log 2x
Let
Taking log 2x as first function and x as second function and integrating by parts, we obtain

Question 6:
x2log x
Let
Taking log x as first function and x2 as second function and integrating by parts, we obtain

Question 7:

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

Question 8:

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

Question 9:

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

Question 10:

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

Question 11:

Let

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

Question 12:

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

Question 13:

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

Question 14:

Taking

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

I=log x 2∫xdx-∫ddxlog x 2∫xdxdx=x22log x 2-∫2log x .1x.x22dx=x22log x 2-∫xlog x dx

Again integrating by parts, we obtain

I = x22logx 2-log x ∫x dx-∫ddxlog x ∫x dxdx=x22logx 2-x22log x -∫1x.x22dx=x22logx 2-x22log x +12∫x dx=x22logx 2-x22log x +x24+C

Question 15:

let
Let I = I1 + I2 … (1)
Where, and

Taking log x as first function and xas 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

Question 16:

Let
Let f(x)=sin x
⇒f’(x)=cos x

It is known that,

Question 17:

Let

Let

It is known that,

Question 18:

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

Question 19:

Also, let
It is known that,

Question 20:

Let
It is known that,

Question 21:

Let
Integrating by parts, we obtain

Again integrating by parts, we obtain

Question 22:

Let
=2θ

Integrating by parts, we obtain

Question 23:
equals

(A)(B) (C) (D)
Let
Also, let

Hence, the correct answer is A.

Question 24:
equals
(A) (B) (C)(D)

Let
Also, let
It is known that,

Hence, the correct answer is B.

### NCERT Solutions For Class 12 Maths Chapter 7 Integrals Ex 7.7 Definite Integral

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

Question 6:

Question 7:

Question 8:

Question 9:

Question 10:
is equal to
A.
B.
C.
D.

Hence, the correct answer is A.

Question 11:
is equal to
A.
B.
C.
D.

Hence, the correct answer is D.

### NCERT Solutions For Class 12 Maths Chapter 7 Integrals Ex 7.8 Fundamental Theorem of Calculus

Question 1:

It is known that,

Question 2:

Let I=
It is known that,

Question 3:

It is known that,

Question 4:

It is known that,

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

Question 5:

Let I= ……….(1)

It is known that,

Question 6:

It is known that,

### NCERT Solutions For Class 12 Maths Chapter 7 Integrals EX 7.9 Evaluation of Definite Integrals by Substitution

Question 1:

By the second fundamental theorem of calculus, we obtain

Question 2:

By the second fundamental theorem of calculus, we obtain

Question 3:

By the second fundamental theorem of calculus, we obtain

Question 4:

By the second fundamental theorem of calculus, we obtain

Question 5:

By the second fundamental theorem of calculus, we obtain

Question 6:

By the second fundamental theorem of calculus, we obtain

Question 7:

By the second fundamental theorem of calculus, we obtain

Question 8:

By the second fundamental theorem of calculus, we obtain

Question 9:

By the second fundamental theorem of calculus, we obtain

Question 10:

By the second fundamental theorem of calculus, we obtain

Question 11:

By the second fundamental theorem of calculus, we obtain

Question 12:

By the second fundamental theorem of calculus, we obtain

Question 13:

By the second fundamental theorem of calculus, we obtain

Question 14:

By the second fundamental theorem of calculus, we obtain

Question 15:

By the second fundamental theorem of calculus, we obtain

Question 16:

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

Question 17:

By the second fundamental theorem of calculus, we obtain

Question 18:

By the second fundamental theorem of calculus, we obtain

Question 19:

By the second fundamental theorem of calculus, we obtain

Question 20:

By the second fundamental theorem of calculus, we obtain

Question 21:
equals
A.
B.
C.
D.

By the second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

Question 22:
equals
A.
B.
C.
D.

By the second fundamental theorem of calculus, we obtain

Hence, the correct answer is C.

### NCERT Solutions For Class 12 Maths Chapter 7 Integrals Ex 7.10 Some Properties of Definite Integrals

Question 1:

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

Question 2:

Also, let

Question 3:

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

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

Question 4:

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

Question 5:

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

Question 6:

Let ⇒ dx = dt

Question 7:

Let x + 1 = ⇒ dx = dt

When x = −1, = 0 and when x = 1, = 2

Question 8:

Let 2x = t ⇒ 2dx = dt

When x = 1, t = 2 and when x = 2, t = 4

Question 9:
The value of the integral is

A. 6

B. 0

C. 3

D. 4

Let cotθ = t ⇒ −cosec2θ dθdt

Hence, the correct answer is A.

Question 10:
If

A. cos x + x sin x

B. x sin x

C. x cos x

D. sin x cos x

Integrating by parts, we obtain

Hence, the correct answer is B.

### NCERT Solutions For Class 12 Maths Chapter 7 Integrals Ex 7.11

Question 1:

Adding (1) and (2), we obtain

Question 2:

Adding (1) and (2), we obtain

Question 3:

Adding (1) and (2), we obtain

Question 4:

Adding (1) and (2), we obtain

Question 5:

Let I=
It can be seen that (x + 2) ≤ 0 on [−5, −2] and (x + 2) ≥ 0 on [−2, 5].

Question 6:

Let I=
It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8].

Question 7:

Question 8:

Question 9:

Question 10:

Adding (1) and (2), we obtain

Question 11:

Let I =

As sin(−x) = (sin (−x))2 = (−sin x)2 = sin2x, therefore, sin2is an even function.

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

Question 12:

Adding (1) and (2), we obtain

Question 13:

Let I= ………(1)
As sin(−x) = (sin (−x))7 = (−sin x)7 = −sin7x, therefore, sin2is an odd function
It is known that, if f(x) is an odd function, then

Question 14:

It is known that,

Question 15:

Adding (1) and (2), we obtain

Question 16:

Adding (1) and (2), we obtain

sin (π − x) = sin x

Adding (4) and (5), we obtain

Let 2x = t ⇒ 2dx = dt

When x = 0, = 0 and when

x=π2, t=π∴

I=12∫0πlog sin tdt-π2log 2

⇒I=I2-π2log 2       [from 3]

⇒I2=-π2log 2

⇒I=-πlog 2

Question 17:

Let I = ………(1)
It is known that,

Adding (1) and (2), we obtain

Question 18:

I =
It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4

Question 19:
Show that if f and g are defined as and

Adding (1) and (2), we obtain

Question 20:
The value of is

A. 0

B. 2

C. π

D. 1

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

Hence, the correct answer is C.

Question 21:
The value of is
A. 2
B.
C.
0
D. -2

Adding (1) and (2), we obtain

Hence, the correct answer is C.

### NCERT Solutions For Class 12 Matas Chapter 7 Integrals Miscellaneous Solutions

Question 1:

Equating the coefficients of x2x, and constant term, we obtain

A + B − C = 0

B + = 0

A = 1

On solving these equations, we obtain

From equation (1), we obtain

Question 2:

Question 3:

Question 4:

Question 5:

On dividing, we obtain

Question 6:

Equating the coefficients of x2x, and constant term, we obtain

A + B = 0

C = 5

9A + = 0

On solving these equations, we obtain

From equation (1), we obtain

Question 7:

Let  a ⇒ dx = dt

Question 8:

Question 9:

Let sin x = t ⇒ cos x dx = dt

Question 10:

Question 11:

Question 12:

Let x= t ⇒ 4x3 dx = dt

Question 13:

Let ex = t ⇒ ex dx = dt

Question 14:

Equating the coefficients of x3x2x, and constant term, we obtain

A + C = 0

B + D = 0

4A + C = 0

4D = 1

On solving these equations, we obtain

From equation (1), we obtain

Question 15:

= cos3 x × sin x

Let cos x = t ⇒ −sin x dx = dt

Question 16:

Question 17:

Question 18:

Question 19:

Let I=∫sin-1x-cos-1xsin-1x+cos-1xdx

It is known that, sin-1x+cos-1x=π2

⇒I=∫π2-cos-1x-cos-1xπ2dx

=2π∫π2-2cos-1xdx

=2π.π2∫1.dx-4π∫cos-1xdx

=x-4π∫cos-1xdx           …(1)

Let I1=∫cos-1x dx

Also, let x=t⇒dx=2 t dt

⇒I1=2∫cos-1t.t dt

=2cos-1t.t22-∫-11-t2.t22dt

=t2cos-1t+∫t21-t2dt

=t2cos-1t-∫1-t2-11-t2dt

=t2cos-1t-∫1-t2dt+∫11-t2dt

=t2cos-1t-t21-t2-12sin-1t+sin-1t

=t2cos-1t-t21-t2+12sin-1t

From equation (1), we obtain

I=x-4πt2cos-1t-t21-t2+12sin-1t  =x-4πxcos-1x-x21-x+12sin-1x

=x-4πxπ2-sin-1x-x-x22+12sin-1x

Question 20:

Question 21:

Question 22:

Equating the coefficients of x2x,and constant term, we obtain

A + C = 1

3A + B + 2= 1

2A + 2B + C = 1

On solving these equations, we obtain

A = −2, B = 1, and C = 3

From equation (1), we obtain

Question 23:

Question 24:

Integrating by parts, we obtain

Question 25:

Question 26:

When = 0, = 0 and

Question 27:

when x=0, t=0 and when

Question 28:

when and when

As therefore

is an even function.

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

Question 29:

Question 30:

Question 31:

From equation (1), we obtain

Question 32:

Adding (1) and (2), we obtain

Question 33:

From equations (1), (2), (3), and (4), we obtain

Question 34:

Equating the coefficients of x2x, 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.

Question 35:

Integrating by parts, we obtain

Hence, the given result is proved.

Question 36:

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.

Question 37:

Hence, the given result is proved.

Question 38:

Hence, the given result is proved.

Question 39:

Integrating by parts, we obtain

Let 1 − x2 = t ⇒ −2x dx = dt

Hence, the given result is proved.

Question 40:
Evaluate as a limit of a sum.
Let I =
It is known that,

Question 41:
is equal to
A.
B.
C.
D.

Hence, the correct answer is A.

Question 42:
is equal to
A.
B.
C.
D.

Hence, the correct answer is B.

Question 43:
If f(a+b-x)=f(x), then is equal to
A.
B.
C.
D.

Hence, the correct answer is D.

Question 44:
The value of is

A. 1

B. 0

C. − 1
D.