**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

**Answer:**

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

It is known that,

Therefore, the antiderivative of sin 2x is

**Question 2:
**Cos 3x

**Answer:**

The antiderivative of cos 3

*x*is a function of

*x*whose derivative is cos 3

*x*

It is known that,

Therefore, the antiderivative of cos 3x is

**Question 3:
**e

^{2}

^{x }

**Answer:**

The antiderivative of *e*^{2}^{x }is the function of* x* whose derivative is *e*^{2}^{x}.

It is known that,

Therefore, the antiderivative of

**Question 4:
Answer:
**The antiderivative of

is the function of

*x*whose derivative is

It is known that,

Therefore, the antiderivative of is

**Question 5:
Answer:
**The antiderivative of is the function of

*x*whose derivative is

It is known that,

Therefore, the antiderivative of is

**Question 6:
Answer:
**

**Question 7:
Answer:
**

**Question 8:
Answer:
**

**Question 9:
Answer:
**

**Question 10:
Answer:
**

**Question 11:
Answer:
**

**Question 12:
Answer:
**

**Question 13:
Answer:
**On dividing, we obtain

**Question 14:
Answer:
**

**Question 15:
Answer:
**

**Question 16:
Answer:
**

**Question 17:
Answer:
**

**Question 18:
Answer:
**

**Question 19:
Answer:
**

**Question 20:
Answer:
**

**Question 21:
**The antiderivative of equals

(A)(B) (C)(D)

**Answer:**

Hence, the correct answer is C.

**Question 22:
**If such that

*f*(2) = 0, then

*f*(

*x*) is

(A)(B)(C)(D)

**Answer:**

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:**

**Answer:
**Let =t

∴2x dx = dt

**Question 2:
**

**Answer:**

Let log |

*x*| =

*t*

**Question 3:
Answer:
**Let 1 + log

*x*=

*t*

∴

∴

**Question 4:
**Sin x.Sin (Cos x)

**Answer:**

in *x* ⋅ sin (cos *x*)

Let cos *x* = *t
∴ −sin x dx = dt
*

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

**Answer:**

let 2(ax+b)=t

∴ 2adx = dt

**Question 6:
Answer:
**

Let *ax + b = t*

⇒ *adx = dt
*

**Question 7:
**

**Answer:**

let (x+2)=t

∴ dx = dt

**Question 8:
**

**Answer:**

Let 1 + 2*x*^{2} = *t*

∴ 4xdx = dt

**Question 9:
Answer:
**Let

∴ (2

*x*+ 1)dx = dt

**Question 10:
Answer:
**let

∴

**Question 11:
Answer:
**Let x+4=t

⇒dx=dt,x=t-4

**Question 12:
Answer:
**Let

∴

**Question 13:
Answer:
**Let

∴ 9

*x*

^{2}dx = dt

**Question 14:
**

**Answer:**

Let log

*x =*t

∴

**Question 15:
Answer:
**Let

∴ −8

*x*dx = dt

**Question 16:
Answer:
**Let 2x+3=t

**∴ 2dx = dt**

**Question 17:
Answer:
**Let

∴ 2xdx = dt

**Question 18:
Answer:
**Let

∴

**Question 19:
**

**Answer:**

Dividing the numerator and denominator by

*e*

^{x}, we obtain

Let

∴

**Question 20:
**

**Answer:**

Let

∴

**Question 21:
**

**Answer:**

Let 2*x* − 3 = *t*

∴ 2*dx** = dt*

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

**Question 22:
Answer:
**

Let 7 − 4x = t

∴ −4dx = dt

**Question 23:
Answer:
**Let

∴

**Question 24:
Answer:
**Let 3 cos x+ 2 sin x =t

∴(-3 sin x +2 cos x)dx=dt

**Question 25:
Answer:
**Let (1-tan x)=t

∴

**Question 26:
Answer:
**Let

∴

**Question 27:
Answer:
**Let sin 2

*x*=

*t*

∴2 cos 2x dx=dt

∴

**Question 28:
Answer:
**Let 1+sin x=t

∴ cos x dx = dt

**Question 29:
**cot

*x*log sin

*x*

**Answer:**

Let log sin

*x*=

*t*

**Question 30:
Answer:
**

Let 1 + cos *x = t*

∴ −sin x dx = dt

**Question 31:
Answer:
**

Let 1 + cos x = t

∴ −sin x dx = dt

**Question 32:
Answer:
**Let sin

*x*+ cos

*x*=

*t*⇒ (cos

*x*− sin

*x*) dx = dt

**Question 33:
Answer:
**Put cos

*x*− sin

*x*=

*t*⇒ (−sin

*x*− cos

*x*) dx = dt

**Question 34:
Answer:
**

**Question 35:
Answer:
**Let 1 + log

*x*=

*t*

∴

∴

**Question 36:
Answer:
**Let (x+log x)=t

∴

**Question 37:
Answer:
**

Let *x*^{4} = *t*

∴ 4*x*^{3}* dx = dt
*Let

∴

From (1), we obtain

**Question 38:
**equals

(A)(B)(C)(D)

**Answer:**

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

**Answer:**

Hence, the correct answer is B.

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

**Question 1:
Answer:
**

**
Question 2:
**sin 3x cos 4x

**Answer:**

It is known that,

**Question 3:
**cos 2x cos 4x cos 6x

**Answer:**

It is known that,

**Question 4:
Answer:
**Let

**Question 5:
Answer:
**

**
Question 6:
**sin x sin 2x sin 3x

**Answer:**

It is known that,

**Question 7:
**sin 4

*x*sin 8

*x*

**Answer:**

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:
Answer:
**

**Question 9:
Answer:
**

**Question 10:
**sin

^{4}

*x*

**Answer:**

**Question 11:
**cos

^{4}2

*x*

**Answer:**

**Question 12:
Answer:
**

**Question 13:
Answer:
**

**Question 14:
Answer:
**

**Question 15:
Answer:
**

**Question 16:
Answer:
**From equation (1), we obtain

**Question 17:
Answer:
**

**Question 18:
Answer:
**

**Question 19:
Answer:
**

1sinxcos3x=sin2x+cos2xsinxcos3x=sinxcos3x+1sinxcosx

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

**Question 20:
**

**Answer:**

**Question 21:
**sin

^{−1}(cos

*x*)

**Answer:**

It is known that,

Substituting in equation (1), we obtain

**Question 22:
**

**Answer:**

**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

**Answer:
**Hence, the correct answer is A.

**Question 24:
**equals

**A.** âˆ’ cot (*e*^{x}*x*) + C

**B.** tan (*xe*^{x}) + C

**C.** tan (*e*^{x}) + C

**D. **cot (*e*^{x}) + C

**Answer:
**Let

*e*

^{x}

*x*=

*t*

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

**Question 1:
Answer:
**

Let *x*^{3} = *t*

∴ 3*x*^{2} *dx* = *dt
*

**Question 2:
Answer:
**

Let 2*x* = *t*

∴ 2*dx* = *dt
*

**Question 3:
Answer:
**

Let 2 − *x *= *t*

⇒ −*dx* = *dt
*

**Question 4:
Answer:
**

Let 5*x* =* t*

∴ 5*dx* = *dt
*

**Question 5:
Answer:
**

Question 6:

Answer:

Let *x*^{3} = *t*

∴ 3*x*^{2} *dx* = *dt
*

**Question 7:
Answer:
**From (1), we obtain

**Question 8:
Answer:
**

Let *x*^{3} = *t*

⇒ 3*x*^{2} *dx* = *dt
*

**Question 9:
Answer:
**

Let tan *x* =* t*

∴ sec^{2}*x* *dx* = *dt
*

**Question 10:
Answer:
**

**Question 11:
**19×2+6x+5

**Answer:**

∫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:
Answer:
**

**Question 13:
Answer:
**

**Question 14:
Answer:
**

**Question 15:
Answer:
**

**Question 16:
Answer:
**

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

4*A* = 4 ⇒ *A* = 1

*A* + *B* = 1 ⇒ *B* = 0

Let 2*x*^{2} + *x* − 3 = *t*

∴ (4*x* + 1) *dx *= *dt
*

**Question 17:
**

**Answer:**

Equating the coefficients of

*x*and constant term on both sides, we obtain

From (1), we obtain

From equation (2), we obtain

**Question 18:
Answer:
**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:
Answer:
**

Equating the coefficients of *x* and constant term, we obtain

2*A* = 6 ⇒ *A* = 3

−9*A* + *B* = 7 ⇒ *B* = 34

∴ 6*x* + 7 = 3 (2*x* − 9) + 34

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

**Question 20:
**

**Answer:**

Equating the coefficients of

*x*and constant term on both sides, we obtain

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

**Question 21:
**

**Answer:**

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

⇒ (2*x* + 2) *dx* =*dt
*Using equations (2) and (3) in (1), we obtain

**Question 22:
**

**Answer:**

Equating the coefficients of

*x*and constant term on both sides, we obtain

*………(3)*

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

**Question 23:
**

**Answer:**

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

**Answer:
**Hence, the correct answer is B.

**Question 25:
**equals

A.

B.

C.

D.

**Answer:**

**Hence, the correct answer is B.**

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

**Question 1:
Answer:
**Let

⇒x=A(x+2)+B(x+1)

Equating the coefficients of *x* and constant term, we obtain

*A* + *B *= 1

2*A* +* B *= 0

On solving, we obtain

*A* = −1 and *B* = 2

**Question 2:
Answer:
**Let

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

Equating the coefficients of *x* and constant term, we obtain

*A* +* B *= 0

−3*A* + 3*B* = 1

On solving, we obtain

**Question 3:
Answer:
**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:
Answer:
**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:
Answer:
**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:
Answer:
**

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

Therefore, on dividing (1 − *x*^{2}) by *x*(1 − 2*x*), we obtain

Substituting *x* = 0 and

in equation (1), we obtain

*A *= 2 and* B *= 3

Substituting equation (1), we obtain

**Question 7:
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

**Question 8:
Answer:
**Let

Substituting

*x*= 1, we obtain

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

*A* + *C* = 0

−2*A* + 2*B* + *C* = 0

On solving, we obtain

**Question 9:
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* − 2*C* = 3

On solving, we obtain

**Question 10:
Answer:
**Let

**Question 11:
Answer:
**Let

Substituting

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

**Question 12:
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

2x+1=A(x-1)+B(x+1)……..(1)

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

**Question 13:
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

**Question 14:
Answer:
**

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

*A* = 3

2*A* + *B *= −1 ⇒ *B* = −7

**Question 15:
**

**Answer:**

Equating the coefficient of

*x*

^{3},

*x*

^{2},

*x*, and constant term, we obtain

By solving these equations, we obtain

**Question 16:
**[Hint: multiply numerator and denominator by

*x*

^{n}

^{ − 1}and put

*x*

^{n}=

*t*]

**Answer:**

Multiplying the numerator and denominator by

*x*

^{n }

^{− 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:
Answer:
**

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

*A* + *C* = 0

*B* + *D* = 4

4*A* + 3*C* = 0

4*B* + 3*D* = 10

On solving these equations, we obtain

*A* = 0, *B* = −2, *C* = 0, and *D* = 6

**Question 19:
Answer:
**Let

*x*

^{2}=

*t*⇒ 2xdx = dt

Substituting

*t*= −3 and

*t*= −1 in equation (1), we obtain

**Question 20:
**

**Answer:**

Multiplying the numerator and denominator by

*x*

^{3}, we obtain

Let

*x*

^{4}=

*t*⇒ 4

*x*

^{3}

*dx*=

*dt*

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

*A* = −1 and *B* = 1

**Question 21:
**[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

**Question 22:
**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.

**Question 23:
**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.

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

**Question 1:
**x sin x

**Answer:**

Let l=

Taking

*x*as first function and sin

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

**Question 2:
**x sin 3x

**Answer:**

Let l=

**Taking**

*x*as first function and sin 3

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

**Question 3:
**

**Answer:**

Let

Taking

*x*

^{2}as first function and

*e*

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

Again integrating by parts, we obtain

**Question 4:
**x log

*x*

**Answer:**

let

Taking log

*x*as first function and

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

**Question 5:
**

*x*log 2

*x*

**Answer:**

Let

Taking log 2

*x*as first function and

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

**Question 6:
**

*x*

^{2}log

*x*

**Answer:**

Let

Taking log

*x*as first function and

*x*

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

**Question 7:
**

**Answer:**

Let

Taking as first function and

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

**
Question 8:
Answer:
**Let

Taking as first function and

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

**Question 9:
Answer:
**Let

Taking cos

^{−1 }

*x*as first function and

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

**Question 10:
Answer:
**Let

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

**Question 11:
Answer:
**Let

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

**Question 12:
**

**Answer:**

Let

Taking

*x*as first function and sec

^{2}

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

**Question 13:**

**Answer:
**Let

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

**Question 14:
Answer:
**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:
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

**Question 16:
**

**Answer:**

Let

Let f(x)=sin x

⇒f’(x)=cos x

∴

It is known that,

∴

**Question 17:
**

**Answer:**

Let

Let

It is known that,

**Question 18:**

**Answer:
**Let

It is known that,

From equation (1), we obtain

**Question 19:
Answer:
**Also, let

It is known that,

**Question 20:
Answer:
**Let

It is known that,

**Question 21:
Answer:
**Let

Integrating by parts, we obtain

Again integrating by parts, we obtain

**Question 22:
Answer:
**Let

=2θ

⇒

Integrating by parts, we obtain

**Question 23:
**equals

(A)(B) (C) (D)

**Answer:
**Let

Also, let

Hence, the correct answer is A.

**Question 24:
**equals

(A) (B) (C)(D)

**Answer:**

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:
Answer:
**

Question 2:

Answer:

Question 3:

Answer:

Question 4:

Answer:

Question 5:

Answer:

Question 6:

Answer:

Question 7:

Answer:

Question 8:

Answer:

Question 9:

Answer:

**
Question 10:
**is equal to

A.

B.

C.

D.

**Answer:**

Hence, the correct answer is A.

**Question 11:
**is equal to

A.

B.

C.

D.

**Answer:**

Hence, the correct answer is D.

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

**Question 1:
Answer:
**

It is known that,

**Question 2:
**

**Answer:**

Let I=

It is known that,

**Question 3:
**

**Answer:**

It is known that,

**Question 4:
**

**Answer:**

It is known that,

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

**Question 5:
Answer:
**Let I= ……….(1)

It is known that,

**Question 6:
Answer:
**

It is known that,

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

**Question 1:
Answer:
**

By the second fundamental theorem of calculus, we obtain

**Question 2:
Answer:
**By the second fundamental theorem of calculus, we obtain

Question 3:

Answer:

By the second fundamental theorem of calculus, we obtain

**Question 4:
Answer:
**By the second fundamental theorem of calculus, we obtain

**Question 5:
Answer:
**

By the second fundamental theorem of calculus, we obtain

**Question 6:
Answer:
**By the second fundamental theorem of calculus, we obtain

**Question 7:
**

**Answer:**

By the second fundamental theorem of calculus, we obtain

**Question 8:
**

**Answer:**

By the second fundamental theorem of calculus, we obtain

**Question 9:
**

**Answer:**

By the second fundamental theorem of calculus, we obtain

**Question 10:
**

**Answer:**

By the second fundamental theorem of calculus, we obtain

**Question 11:
Answer:
**By the second fundamental theorem of calculus, we obtain

**Question 12:
**

**Answer:**

By the second fundamental theorem of calculus, we obtain

**Question 13:
Answer:
**By the second fundamental theorem of calculus, we obtain

**Question 14:
Answer:
**By the second fundamental theorem of calculus, we obtain

**Question 15:
Answer:
**

By the second fundamental theorem of calculus, we obtain

**Question 16:
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

**Question 17:
**

**Answer:**

By the second fundamental theorem of calculus, we obtain

**Question 18:
**

**Answer:**

By the second fundamental theorem of calculus, we obtain

**Question 19:
Answer:
**By the second fundamental theorem of calculus, we obtain

**Question 20:
**

**Answer:**

By the second fundamental theorem of calculus, we obtain

**Question 21:
**equals

**Answer:**

A.

B.

C.

D.

**Answer:**

By the second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

**Question 22:
**equals

**Answer:**

A.

B.

C.

D.

**Answer:**

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:
**

**Answer:**

When

*x*= 0,

*t*= 1 and when

*x*= 1,

*t*= 2

**Question 2:
Answer:
**Also, let

**Question 3:
**

**Answer:**

Let I=

Also, let

*x*= tan

*θ*⇒

*dx*= sec

^{2}

*θ*d

*θ*

When

*x*= 0,

*θ*= 0 and when

*x*= 1, θ=

Taking

*θ*as the first function and sec

^{2}

*θ*as the second function and integrating by parts, we obtain

**Question 4:
**

**Answer:**

Let

*x*+ 2 =

*t*

^{2}⇒

*dx*= 2

*tdt*

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

**Question 5:
**

**Answer:**

Let cos

*x*=

*t*⇒ −sin

*x*

*dx*=

*dt*

When x = 0, t = 1 and when

**Question 6:
**

**Answer:**

Let ⇒ dx = dt

**Question 7:
**

**Answer:**

Let *x* + 1 = *t *⇒ *dx* = *dt*

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

**Question 8:
Answer:
**

Let 2*x* =* t* ⇒ 2*dx* = *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

**Answer:
**Let cot

*θ*=

*t*⇒ −cosec2

*θ*

*d*

*θ*=

*dt*

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 *+ *x* cos *x
*

**Answer:**

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

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

**Question 3:
**

**Answer:**

Adding (1) and (2), we obtain

**Question 4:
**

**Answer:**

Adding (1) and (2), we obtain

**Question 5:
Answer:
**Let I=

It can be seen that (

*x*+ 2) ≤ 0 on [−5, −2] and (

*x*+ 2) ≥ 0 on [−2, 5].

**Question 6:
Answer:
**Let I=

It can be seen that (

*x*− 5) ≤ 0 on [2, 5] and (

*x*− 5) ≥ 0 on [5, 8].

**Question 7:
Answer:
**

**Question 8:
Answer:
**

**Question 9:
Answer:
**

**Question 10:
Answer:
**

Adding (1) and (2), we obtain

**Question 11:
Answer:
**Let I =

As sin^{2 }(−*x*) = (sin (−*x*))^{2} = (−sin *x*)^{2} = sin^{2}*x*, therefore, sin^{2}*x *is an even function.

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

**Question 12:
Answer:
**Adding (1) and (2), we obtain

**Question 13:
Answer:
**Let I= ………(1)

As sin

^{7 }(−

*x*) = (sin (−

*x*))

^{7}= (−sin

*x*)

^{7}= −sin

^{7}

*x*, therefore, sin

^{2}

*x*is an odd function

It is known that, if

*f*(

*x*) is an odd function, then

**Question 14:
Answer:
**It is known that,

**Question 15:
Answer:
**Adding (1) and (2), we obtain

**Question 16:
**

**Answer:**

Adding (1) and (2), we obtain

sin (π −

*x*) = sin

*x*

Adding (4) and (5), we obtain

Let 2*x* = *t* ⇒ 2*dx* = *dt*

When *x* = 0, *t *= 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:
Answer:
**Let I = ………(1)

It is known that,

Adding (1) and (2), we obtain

**Question 18:
Answer:
**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

**Answer:**

Adding (1) and (2), we obtain

**Question 20:
**The value of is

**A. **0

**B. **2

**C. **π

**D.** 1

**Answer:
**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

C.

**D.**-2

**Answer:**

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:
**

**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

**Question 2:
Answer:
**

**Question 3:
**[Hint: Put ]

**Answer:**

**Question 4:
Answer:
**

**Question 5:
Answer:
**On dividing, we obtain

**Question 6:
**

**Answer:**

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

*A* + *B* = 0

*B *+ *C* = 5

9*A* + *C *= 0

On solving these equations, we obtain

From equation (1), we obtain

**Question 7:
Answer:
**Let

*x*−

*a*=

*t*⇒

*dx*=

*dt*

**Question 8:
**

**Answer:**

**Question 9:**

**Answer:
**Let sin

*x*=

*t*⇒ cos

*x dx*=

*dt*

**Question 10:
**

**Answer:**

**Question 11:
**

**Answer:**

**Question 12:
**

**Answer:**

Let

*x*

^{4 }=

*t*⇒ 4

*x*

^{3}

*dx*=

*dt*

**Question 13:
**

**Answer:**

Let

*e*

^{x}=

*t*⇒

*e*

^{x}

*dx*=

*dt*

**Question 14:
**

**Answer:**

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

*A* + *C* = 0

*B* + *D* = 0

4*A* + *C* = 0

4*B *+ *D* = 1

On solving these equations, we obtain

From equation (1), we obtain

**Question 15:
**

**Answer:**

= cos^{3} *x* × sin *x*

Let cos *x* =* t* ⇒ −sin *x dx* =* dt
*

**Question 16:
Answer:
**

**Question 17:
Answer:
**

**Question 18:
Answer:
**

**Question 19:
Answer:
**

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:
Answer:
**

**Question 21:
Answer:
**

**Question 22:
Answer:
**

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

*A* + *C* = 1

3*A* + *B* + 2*C *= 1

2*A* + 2*B* + *C* = 1

On solving these equations, we obtain

*A* = −2, *B* = 1, and *C* = 3

From equation (1), we obtain

**Question 23:
Answer:
**

**Question 24:
Answer:
**Integrating by parts, we obtain

**Question 25:
**

**Answer:**

**Question 26:
**

**Answer:**

**When**

*x*= 0,

*t*= 0 and

**Question 27:
Answer:
**when x=0, t=0 and when

**Question 28:
Answer:
**when and when

As therefore

is an even function.

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

**Question 29:
Answer:
**

**Question 30:
Answer:
**

**Question 31:
Answer:
**From equation (1), we obtain

**Question 32:
Answer:
**Adding (1) and (2), we obtain

**Question 33:
Answer:
**

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

**Question 34:
**

**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.

**Question 35:
**

**Answer:**

Integrating by parts, we obtain

Hence, the given result is proved.

**Question 36:
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.

**Question 37:
**

**Answer:**

Hence, the given result is proved.

**Question 38:
**

**Answer:**

Hence, the given result is proved.

**Question 39:
**

**Answer:**

Integrating by parts, we obtain

Let 1 − *x*^{2} = *t* ⇒ −2*x* *dx* = *dt*

Hence, the given result is proved.

**Question 40:
**Evaluate as a limit of a sum.

**Answer:**

Let I =

It is known that,

**Question 41:
**is equal to

A.

B.

C.

D.

**Answer:**

Hence, the correct answer is A.

**Question 42:**

is equal to

A.

B.

C.

D.

**Answer:
**Hence, the correct answer is B.

**Question 43:
**If f(a+b-x)=f(x), then is equal to

A.

B.

C.

D.

**Answer:**

Hence, the correct answer is D.

**Question 44:
**The value of is

**A.** 1

**B.** 0

**C.** − 1

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

Hence, the correct answer is B.