# NCERT Solutions for Class 12 Maths Part 2 Chapter 7 Integrals

Free NCERT Solutions for Class 12 Maths Chapter 7 Integrals solved by Expert Teachers as per NCERT (CBSE) Book guidelines and brought to you by Toppers Bulletin. These Integrals Exercise Questions with Solutions for Class 12 Maths covers all questions of Chapter Integrals Class 12 and help you to revise complete Syllabus and Score More marks as per CBSE Board guidelines from the latest NCERT book for class 12 maths. You can read and download NCERT Book Solution to get a better understanding of all topics and concepts

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

**Exercise 7.1 :** Solutions of Questions on Page Number** : 299**

**Q1 :sin 2x**

**Answer :**The anti derivative of sin 2*x* is a function of *x* whose derivative is sin 2*x*.

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 :**

**Answer :**

The anti derivative of is the function of x whose derivative is

It is known that,

Therefore, the anti derivative of is .

**Q6 :**

**Answer :**

**Q7 :**

**Answer :**

**Q8 :**

**Answer :**

**Q9 :**

**Answer :**

**Q10 :**

**Answer :**

**Q11 :**

**Answer :**

**Q12 :**

**Answer :**

**Q13 :**

**Answer :**

On dividing, we obtain

**Q14 :**

**Answer :**

**Q15 :**

**Answer :**

**Q16 :**

**Answer :**

**Q17 :**

**Answer :**

**Q18 :**

**Answer :**

**Q19 :**

**Answer :**

**Q20 :**

**Answer :**

**Q21 :The anti derivative of equals**

**(A) (B)**

**(C) (D)**

**Answer :**

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 :**

**Answer :**

Let = t

∴2x dx = dt

**Q2 :**

**Answer :**

Let log |x| = t

∴

**Q3 :**

**Answer :**

Let 1 + log x = t

∴

**Q4 :sin x . sin (cos x)
**

**Answer :**sin x Ã¢â€¹â€¦ sin (cos x)

Let cos x = t

∴ – sin x dx = dt

**Q5 :**

**Answer :**

Let∴ 2adx = dt

**Q6 :**

**Answer :
**Let ax + b = t

⇒ adx = dt

**Q7 :**

**Answer :**Let

∴ dx = dt

**Q8 :**

**Answer :
**Let 1 + 2×2 = t

∴ 4xdx = dt

**Q9 :**

**Answer :
**Let

∴ (2x + 1)dx = dt

**Q10 :**

**Answer : **

Let

∴

**Q11 :**

**Answer :
**

**Q12 :**

**Answer :
**Let

∴

**Q13 :**

**Answer :
**Let

∴ 9×2 dx = dt

**Q14 :**

**Answer :
**Let log x = t

∴

**Q15 :**

**Answer :
**Let∴ – 8x dx = dt

**Q16 :**

**Answer :
**Let∴ 2dx = dt

**Q17 :**

**Answer :
**Let∴ 2xdx = dt

**Q18 :**

**Answer :
**Let

**Q19 :**

**Answer :**

Dividing numerator and denominator by ex, we obtain

Let

∴

**Q20 :**

**Answer :
**Let

∴

**Q21 :**

**Answer :**

**
**Let 2x – 3 = t

∴ 2dx = dt

**Q22 :**

**Answer :
**Let 7 – 4x = t

∴ – 4dx = dt

**Q23 :**

**Answer :
**Let

∴

**Q24 :**

**Answer :**

Let

∴

**Q25 :**

**Answer :
**Let

∴

**Q26 :**

**Answer :
**Let

∴

**Q27 :**

**Answer :
**Let sin 2x = t

∴

**Q28 :**

**Answer :
**Let∴ cos x dx = dt

**Q29 :****cot x log sin x**

**Answer :
**Let log sin x = t

**Q30 :**

**Answer :
**Let 1 + cos x = t

∴ – sin x dx = dt

**Q31 :**

**Answer :
**Let 1 + cos x = t

∴ – sin x dx = dt

**Q32 :**

**Answer :**

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

**Q33 :**

**Answer :**

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

**Q34 :**

**Answer :**

**Q35 :**

**Answer :
**Let 1 + log x = t

∴

**Q36 :**

**Answer :**

Let

∴

**Q37 :**

**Answer :
**Let x4 = t

∴ 4x3dx = dt

Let

∴

From (1), we obtain

**Q38 :****equals**

**Answer :
**Let

∴

Hence, the correct answer is D.

**Q39 :****equals**

**Answer :
**Let

∴

Hence, the correct answer is D.

**Exercise 7.3 : **Solutions of Questions on Page Number **: 307**

**Q1 :**

**Answer :**

**Q2 :**

**Answer :
**It is known that,

**Q3 : ****cos 2x cos 4x cos 6x**

**Answer :
**It is known that,

**Q4 : ****sin3 (2x + 1)**

**Answer :
**Let

**Q5 : ****sin3 x cos3 x**

**Answer :**

**Q6 : ****sin x sin 2x sin 3x**

**Answer :
**It is known that,

**Q7 : ****sin 4x sin 8x**

**Answer :
**It is known that,

**Q8 :**

**Answer :**

**Q9 :**

**Answer :**

**Q10 : ****sin4 x**

**Answer :**

**Q11 : ****cos4 2x**

**Answer :**

**Q12 :**

**Answer :**

**Q13 :**

**Answer :**

**Q14 :**

**Answer :**

**Q15 :**

**Answer :**

**Q16 : ****tan4x**

**Answer :**

From equation (1), we obtain

**Q17 :**

**Answer :**

**Q18 :**

**Answer :**

**Q19 :**

**Answer :
**

**Q20 :**

**Answer :**

**Q21 : ****sin-1 (cos x)**

**Answer :**

It is known that,

Substituting in equation (1), we obtain

**Q22 :**

**Answer :**

**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**

**Answer :**

Hence, the correct answer is A.

**Q24 :****equals**

**A. – cot (exx) + C
**

**B. tan (xex) + C**

**C. tan (ex) + C**

**D. cot (ex) + C**

**Answer :**

Let exx = t

Hence, the correct answer is B.

**Exercise 7.4 : **Solutions of Questions on Page Number** : 315**

**Q1 :**

**Answer :
**Let x

^{3}= t

∴ 3x

^{2}dx = dt

**Q2 :**

**Answer :
**Let 2x = t

∴ 2dx = dt

**Q3 :**

**Answer :
**Let 2 – x = t

⇒ – dx = dt

**Q4 :**

**Answer :
**Let 5x = t

∴ 5dx = dt

**Q5 :**

**Answer :**

**Q6 :**

**Answer :
**Let x3 = t

∴ 3×2 dx = dt

**Q7 :**

**Answer :**

From (1), we obtain

**Q8 :**

**Answer :
**Let x

^{3}= t

⇒ 3x

^{2}dx = dt

**Q9 :**

**Answer :
**Let tan x = t

∴ sec

^{2}x dx = dt

**Q10 :**

**Answer :**

**Q12 :**

**Answer :**

**Q13 :**

**Answer :**

**Q14 :**

**Answer :**

**Q15 :**

**Answer :**

**Q16 :**

**Answer :**

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 :**

**Answer :**

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

From (1), we obtain

From equation (2), we obtain

**Q18 :
Answer:**

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

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

**Q19 :**

**Answer :Answre **

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 :**

**Answer :**

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

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

**Q21 :**

**Answer :**

Let x^{2} + 2x +3 = t

⇒ (2x + 2) dx =dt

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

**Q22 :**

**Answer :**

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

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

**Q23 :**

**Answer :**

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**

**Answer :**

Hence, the correct answer is B.

**Q25 :****equals**

**A.**

**B.**

**C.**

**D.**

**Answer :**

Hence, the correct answer is B.

**Exercise 7.5 :** Solutions of Questions on Page Number** : 322**

**Q1 :**

**Answer :
**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 :**

**Answer :
**Let

Equating the coefficients of x and constant term, we obtain

A + B = 0

– 3A + 3B = 1

On solving, we obtain

**Q3 :**

**Answer :
**Let

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

A = 1, B = – 5, and C = 4

**Q4 :**

**Answer :
**Let

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

**Q5 :**

**Answer :
**Let

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

A = – 2 and B = 4

**Q6 :**

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

Therefore, on dividing (1 – x

^{2}) 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 :**

**Answer :
**Let

Equating the coefficients of x

^{2}, 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 :**

**Answer :
**Let

Substituting x = 1, we obtain

Equating the coefficients of x

^{2}and constant term, we obtain

A + C = 0

– 2A + 2B + C = 0

On solving, we obtain

**Q9 :**

**Answer :**

Let

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

B = 4

Equating the coefficients of x^{2} and x, we obtain

A + C = 0

B – 2C = 3

On solving, we obtain

**Q10 :**

**Answer :**

Let

Equating the coefficients of x2 and x, we obtain

**Q11 :**

**Answer :**

Let

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

**Q12 :**

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

Therefore, on dividing (x

^{3}+ x + 1) by x

^{2}– 1, we obtain

Let

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

**Q13 :**

**Answer :**

Equating the coefficient of x^{2}, 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 :**

**Answer :**

Equating the coefficient of x and constant term, we obtain

A = 3

2A + B = – 1 ⇒ B = – 7

**Q15 :**

**Answer :**

Equating the coefficient of x^{3}, x^{2}, x, and constant term, we obtain

On solving these equations, we obtain

**Q16 :****[Hint: multiply numerator and denominator by x ^{n – 1} and put x^{n} = t]**

**Answer :**

Multiplying numerator and denominator by x

^{n – 1}, we obtain

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

A = 1 and B = – 1

**Q17 :****[Hint: Put sin x = t]**

**Answer :**

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

A = 1 and B = – 1

**Q18 :**

**Answer :**

Equating the coefficients of x^{3}, x^{2}, 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 :**

**Answer :**

Let x^{2} = t ⇒ 2x dx = dt

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

**Q20 :**

**Answer :**

Multiplying numerator and denominator by x^{3}, we obtain

Let x4 = t ⇒ 4x3dx = dt

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

A = – 1 and B = 1

**Q21 :****[Hint: Put e ^{x} = t]**

**Answer :**

Let e

^{x}= t ⇒ e

^{x}dx = dt

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

A = – 1 and B = 1

**Q22 :**

**A.**

**B.**

**C.**

**D.**

**Answer :**

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.**

**Answer :**

Equating the coefficients of x^{2}, 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
**

**Answer :**

Let I =

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

**Q2 :**

**Answer :
**Let I =

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

**Q3 :**

**Answer :
**Let

Taking x

^{2}as first function and ex as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

**Q4 : ****x logx
**

**Answer :**

Let

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

**Q5 :****x log 2x
**

**Answer :**

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

**Q6 : ****x ^{2} log x
**

**Answer :**

LetTaking log x as first function and x

^{2}as second function and integrating by parts, we obtain

**Q7 :**

**Answer :
**Let

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

**Q8 :
**

**Answer :**

Let

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

**Q9 :
**

**Answer :**

Let

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

**Q10 :
**

**Answer :**

Let

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

**Q11 :
**

**Answer :**

Let

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

**Q12 :
**

**Answer :**

Let

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

**Q13 :
**

**Answer :**

Let

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

**Q14 :
**

**Answer :**

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

**Q15 :
**

**Answer :**

Let

Let I = I

_{1}+ I

_{2}… (1)

Where,and

Taking log x as first function and x

^{2}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 :
**

**Answer :**

Let

Let

⇒

∴

It is known that,

**Q17 :
**

**Answer :**

Let

Let

⇒

It is known that,

**Q18 :
**

**Answer :**

Let⇒

It is known that,

From equation (1), we obtain

**Q19 :
**

**Answer :**

Also, let ⇒

It is known that,

**Q20 :
**

**Answer :**

Let

⇒

It is known that,

**Q21:
**

**Answer :**

Let

Integrating by parts, we obtain

Again integrating by parts, we obtain

**Q22 :
**

**Answer :**

Let ⇒

= 2θ

⇒

Integrating by parts, we obtain

**Q23 :****equals**

**Answer :**

Let

Also, let ⇒

Hence, the correct answer is A.

**Q24 :****equals**

**Answer :**

**
**Let

Also, let ⇒

It is known that,

Hence, the correct answer is B.

**Exercise 7.7 :** Solutions of Questions on Page Number** : 330**

**Q1 :**

**Answer :**

**Q2 :**

**Answer :**

**Q3 :**

**Answer :**

**Q4 :**

**Answer :**

**Q5 :**

**Answer :**

**Q6 :**

**Answer :**

**Q7 :**

**Answer :**

**Q8 :**

**Answer :**

**Q9 :**

**Answer :**

**Q10 :****is equal to**

**A.
**

**B.**

**C.**

**D.**

**Answer :**

Hence, the correct answer is A.

**Q11 :****is equal to
**

**A.**

**B.**

**C.**

**D.**

**Answer :**

Hence, the correct answer is D.

**Exercise 7.8 :** Solutions of Questions on Page Number **: 334**

**Q1 :**

**Answer :
**It is known that,

**Q2 :**

**Answer :**

It is known that,

**Q3 :**

**Answer :
**It is known that,

**Q4 :**

**Answer :**

It is known that,

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

**Q5 :**

**Answer :**

It is known that,

**Q6 :**

**Answer :
**It is known that,

**Exercise 7.9 :** Solutions of Questions on Page Number **: 338**

**Q1 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q2 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q3 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q4 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q5 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q6 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q7 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q8 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q9 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q10 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q11 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q12 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q13 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q14 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q15 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q16 :**

**Answer :
**Let

Equating the coefficients of x and constant term, we obtain

A = 10 and B = – 25

Substituting the value of I

_{1}in (1), we obtain

**Q17 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q18 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q19 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q20 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q21 : ****equals
**

**A.**

**B.**

**C.**

**D.**

**Answer :**

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

**Q22 : ****equals
**

**A.**

**B.**

**C.**

**D.**

**Answer :**

**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 :**

**Answer :**

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

**Q2 :**

**Answer :**

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

**Q3 :**

**Answer :**

Also, let

**Q4 :**

**Answer :**

Also, let

**Q5 :**

**Answer :**

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 :**

**Answer :**

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 :**

**Answer :**

Let x + 2 = t^{2} ⇒ dx = 2tdt

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

**Q8 :**

**Answer :**

Let x + 2 = t^{2} ⇒ dx = 2tdt

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

**Q9 :**

**Answer :**

Let cos x = t ⇒ – sinx dx = dt

When x = 0, t = 1 and when

**Q10 :**

**Answer :**

Let cos x = t ⇒ – sinx dx = dt

When x = 0, t = 1 and when

**Q11 :**

**Answer :**

**
**Let ⇒ dx = dt

**Q12 :**

**Answer :**

**
**Let ⇒ dx = dt

**Q13 :**

**Answer :**

Let x + 1 = t ⇒ dx = dt

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

**Q14 :**

**Answer :**

Let x + 1 = t ⇒ dx = dt

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

**Q15 :**

**Answer :**

Let 2x = t ⇒ 2dx = dt

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

**Q16 :**

**Answer :**

** **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**

**Answer :**

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**

**Answer :**

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**

**Answer :**

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**

**Answer :**

Integrating by parts, we obtain

Hence, the correct answer is B.

**Exercise 7.11 :** Solutions of Questions on Page Number** : 347**

**Q1 :**

**Answer :**

Adding (1) and (2), we obtain

**Q2 :**

**Answer :**

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

**Q3 :**

**Answer :**

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

**Q4 :**

**Answer :**

Adding (1) and (2), we obtain

**Q5 :****Answer :**

Adding (1) and (2), we obtain

**Q6 :**

**Answer :
**Adding (1) and (2), we obtain

**Q7 :****Answer :**

Adding (1) and (2), we obtain

**Q8 :** **Answer :**

**Q9 :**

**Answer :**

**Q10 :
**

**Answer :**

Adding (1) and (2), We obtain

**Q11 :** **Answer :
**As sin

^{2}(-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 :**

**Answer :**

**Q13 :**

**Answer :
**As sin

^{7}(-x)=(sin(-x))

^{7}=(-sinx)

^{7}=-sin

^{7}x,therefore,sin

^{2}x is an even function.

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

**Q14 :** **
Answer :
**It is known that,

**Q15 :**

**Answer :
**

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

**Q16 :**

**Answer :
**Adding (1) and (2), we obtain,

sin(Π-x)=sinx

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

**Q17 :**

**Answer :
**It is known that,

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

**Q18 :**

**Answer :
**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
**

**Answer :**

Adding (1) and (2), we obtain

**Q20 : The value of
A**.0

**B**.2

**C.**Π

**D.**

**Answer :**

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

and f(x) is an odd function,then

Hence,the correct answer is C.

**Q21 : The value of
**

**A.**2

**0**

B. C.

B. C.

**-2**

D.

D.

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

Answer :

Answer :

Hence the correct answer is C.

**Exercise Miscellaneous :** Solutions of Questions on Page Number** : 352**

**Q1 :**

**Answer :
**Equating the coefficients of x

^{2}, 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 :**

**Answer :
**

**Q3 :**

** [Hint: Put]**

**Answer :
**

**Q4 :**

**Answer :
**

**Q5 :**

**Answer :
**

On dividing, we obtain

**Q6 :**

**Answer :
**

Equating the coefficients of x

^{2}, 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 :**

**Answer :
**Let x – a = t ⇒ dx = dt

**Q8 :
**

**Answer :**

**Q9 :**

**Answer :
**Let sin x = t ⇒ cos x dx = dt

**Q10 :**

**Answer :
**

**Q11 :**

**Answer :
**

**Q12 :**

**Answer :
**Let x

^{4}=t ⇒ 4x

^{3}dx = dt

**Q13 :**

**Answer :
**Let ex = t ⇒ e

^{x}dx = dt

**Q14 :**

**Answer :
**Equating the coefficients of x

^{3}, x

^{2}, 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 :
**

**Answer :**

= cos3 x × sin x

Let cos x = t ⇒ – sin x dx =dt

**Q16 :**

**Answer :
**

**Q17 :**

**Answer :**

**Q18 :**

**Answer :
**

**Q19 :**

**Answer :
**

**Q20 :**

**Answer :
**

**Q21 :
**

**Answer :**

**Q22 :**

**Answer :
**Equating the coefficients of x

^{2}, 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 :**

**Answer :
**

**Q24 :**

**Answer :
**

Integrating by parts, we obtain

**Q25 :**

**Answer :
**

**Q26 :**

**Answer :
**

When x = 0, t = 0 and

**Q27 :**

**Answer :
**When and when

**Q28 :**

**Answer :
**

When and when

As ,

therefore, is an even function.

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

**Q29 :**

**Answer :**

**Q30 :**

**Answer :
**

**Q31 :**

**Answer :
**

From equation (1), we obtain

**Q32 :**

**Answer :
**

Adding (1) and (2), we obtain

**Q33 :**

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

**Q34 :**

**Answer :
**Equating the coefficients of x

^{2}, 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.

**Q35 :** **Answer :
**Integrating by parts, we obtain

Hence, the given result is proved.

**Q36 :
**

**Answer :**

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 :**

**Answer :
**

Hence, the given result is proved.

**Q38 :****Answer :
**

Hence, the given result is proved.

**Q39 :**

**Answer :
**Integrating by parts, we obtain

Let 1 – x^{2} = t ⇒ – 2x dx = dt

Hence, the given result is proved.

**Q40 :**

** Evaluate as a limit of a sum.**

**Answer :
**It is known that,

**Q41 :**

** is equal to**

**A.**

**B.**

**C.**

**D.**

**Answer :
**

Hence, the correct answer is A.

**Q42 :**

is equal to

**A.**

**B.**

**C.**

**D.**

**Answer :
**

Hence, the correct answer is B.

**Q43 :** **If then is equal to**

**A.**

**B.**

**C.**

**D.**

**Answer :**

Hence, the cor

Correct answer is D.

**Q44 :**

** The value of is**

**A. 1**

**B. 0**

**C. – 1**

**D.**

**Answer :
**Adding (1) and (2), we obtain

Hence, the correct answer is B.