Class 12 Maths NCERT Solutions

NCERT Class 12 Maths Solutions Chapter 7 – Integrals

NCERT Class 12 Maths Solutions Chapter 7 – Integrals: Hello dear students. In this page, you will get the full solution of Integrals. So solve the questions of this chapter at your home.

In our previous post, we have updated solution of Chapter 6 i.e Applications of Derivatives. If you have not checked it yet, then check it now. You can download pdf of this page in your mobile/laptop.

NCERT Class 12 Maths Solutions Chapter 7 – Integrals

The topics and sub-topics included in the Integrals chapter are the following:

Section NameTopic Name
7Integrals
7.1Introduction
7.2Integration as an Inverse Process of Differentiation
7.3Methods of Integration
7.4Integrals of some Particular Functions
7.5Integration by Partial Fractions
7.6Integration by Parts
7.7Definite Integral
7.8Fundamental Theorem of Calculus
7.9Evaluation of Definite Integrals by Substitution
7.10Some Properties of Definite Integrals

NCERT Solutions for Class 12 Maths Chapter 7 Integrals

Page No 299:

Question 1:

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

Question 2:

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 .

Question 3:

e2x

Answer:

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

It is known that,

Therefore, the anti derivative of .

Question 4:

Answer:

The anti derivative of

is the function of whose derivative is .

It is known that,

Therefore, the anti derivative of .

Question 5:

Answer:

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

It is known that,

Therefore, the anti derivative 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 anti derivative 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,

∴Anti derivative of 

Also,

Hence, the correct answer is A.

Page No 304:

Question 1:

Answer:

Let t

∴2x dx = dt

Question 2:

Answer:

Let log |x| = t

∴ 

Question 3:

Answer:

Let 1 + log t

∴ 

Question 4:

sin x ⋅ sin (cos x)

Answer:

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt

Question 5:

Answer:

Let 

∴ 2adx = dt

Question 6:

Answer:

Let ax + b = t

⇒ adx = dt

Question 7:

Answer:

Let 

∴ dx = dt

Question 8:

Answer:

Let 1 + 2x2 = t

∴ 4xdx = dt

Question 9:

Answer:

Let 

∴ (2x + 1)dx = dt

Question 10:

Answer:

Let 

Question 11:

Answer:

Let I=∫xx+4 dxput x+4=t⇒dx=dtNow, I=∫t-4tdt=∫t-4t-1/2dt=23t3/2-42t1/2+C=23.t.t1/2-8t1/2+C=23x+4x+4-8x+4+C=23xx+4+83x+4-8x+4+C=23xx+4-163x+4+C=23x+4x-8+C

Question 12:

Answer:

Let 

∴ 

Question 13:

Answer:

Let 

∴ 9x2 dx = dt

Question 14:

Answer:

Let log x = t

∴ 

Question 15:

Answer:

Let 

∴ −8x dx = dt

Question 16:

Answer:

Let 

∴ 2dx = dt

Question 17:

Answer:

Let 

∴ 2xdx = dt

Page No 305:

Question 18:

Answer:

Let 

∴ 

Question 19:

Answer:

Dividing numerator and denominator by ex, we obtain

Let 

∴ 

Question 20:

Answer:

Let 

∴ 

Question 21:

Answer:

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:

Answer:

Let 7 − 4x = t

∴ −4dx = dt

Question 23:

Answer:

Let 

∴ 

Question 24:

Answer:

Let 

∴ 

Question 25:

Answer:

Let 

∴ 

Question 26:

Answer:

Let 

∴ 

Question 27:

Answer:

Let sin 2x = t

∴ 

Question 28:

Answer:

Let 

∴ 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 xdx = dt

Question 33:

Answer:

Put cos x − sin x = t ⇒ (−sin x − cos xdx = dt

Question 34:

Answer:

Question 35:

Answer:

Let 1 + log x = t

∴ 

Question 36:

Answer:

Let

∴ 

Question 37:

Answer:

Let x4 = t

∴ 4x3 dx = dt

Let 

From (1), we obtain

Question 38:

equals

Answer:

Let 

∴ 

Hence, the correct answer is D.

Question 39:

equals

A. 

B. 

C. 

D. 

Answer:

Hence, the correct answer is B.

Page No 307:

Question 1:

Answer:

Question 2:

Answer:

It is known that, 

Question 3:

cos 2x cos 4x cos 6x

Answer:

It is known that,

Question 4:

sin3 (2x + 1)

Answer:

Let 

Question 5:

sin3 x cos3 x

Answer:

Question 6:

sin x sin 2x sin 3x

Answer:

It is known that, 

Question 7:

sin 4x sin 8x

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:

sin4 x

Answer:

Question 11:

cos4 2x

Answer:

Question 12:

Answer:

Question 13:

Answer:

Question 14:

Answer:

Question 15:

Answer:

Question 16:

tan4x

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 (exx) + C

B. tan (xex) + C

C. tan (ex) + C

D. cot (ex) + C

Answer:

Let exx = t

Hence, the correct answer is B.

Page No 315:

Question 1:

Answer:

Let x3 = t

∴ 3x2 dx = dt

Question 2:

Answer:

Let 2x = t

∴ 2dx = dt

Question 3:

Answer:

Let 2 − t

⇒ −dx = dt

Question 4:

Answer:

Let 5x = t

∴ 5dx = dt

Question 5:

Answer:

Question 6:

Answer:

Let x3 = t

∴ 3x2 dx = dt

Question 7:

Answer:

From (1), we obtain

Question 8:

Answer:

Let x3 = t

⇒ 3x2 dx = dt

Question 9:

Answer:

Let tan x = t

∴ sec2x dx = dt

Page No 316:

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

4A = 4 ⇒ A = 1

A + B = 1 ⇒ B = 0

Let 2x2 + x − 3 = t

∴ (4x + 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

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

Question 19:

Answer:

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:

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 x2 + 2x +3 = t

⇒ (2x + 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

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.

Page No 322:

Question 1:

Answer:

Let 

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:

Answer:

Let 

Equating the coefficients of x and constant term, we obtain

A + B = 0

−3A + 3B = 1

On solving, we obtain

Question 3:

Answer:

Let 

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

A = 1, B = −5, and C = 4

Question 4:

Answer:

Let 

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

Question 5:

Answer:

Let 

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 − x2) by x(1 − 2x), we obtain

Let 

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

= 2 and B = 3

Substituting in equation (1), we obtain

Question 7:

Answer:

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:

Answer:

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:

Answer:

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:

Answer:

Let 

Equating the coefficients of x2 and x, we obtain

Question 11:

Answer:

Let 

Substituting = −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 (x3 + x + 1) by x2 − 1, we obtain

Let 

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

Question 13:

Answer:

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:

Answer:

Equating the coefficient of x and constant term, we obtain

A = 3

2A + = −1 ⇒ B = −7

Question 15:

Answer:

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

On solving these equations, we obtain

Question 16:

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

Answer:

Multiplying 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

Page No 323:

Question 18:

Answer:

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:

Answer:

Let x2 = t ⇒ 2x dx = dt

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

Question 20:

Answer:

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

Question 21:

 [Hint: Put ex = t]

Answer:

Let ex = ⇒ ex 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 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.

Page No 327:

Question 1:

x sin x

Answer:

Let I = 

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

Question 2:

Answer:

Let I = 

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

Question 3:

Answer:

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 logx

Answer:

Let 

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

Question 5:

x log 2x

Answer:

Let 

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

Question 6:

xlog x

Answer:

Let 

Taking log x as first function and x2 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 sec2x 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 = 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

Page No 328:

Question 16:

Answer:

Let 

Let

⇒ 

∴ 

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

Answer:

Let 

Also, let  ⇒ 

Hence, the correct answer is A.

Question 24:

 equals

Answer:

Let 

Also, let  ⇒ 

It is known that, 

Hence, the correct answer is B.

Page No 330:

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.

Page No 334:

Question 1:

Answer:

It is known that,

Question 2:

Answer:

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:

It is known that,

Question 6:

Answer:

It is known that,

Page No 338:

Question 1:

Answer:

By second fundamental theorem of calculus, we obtain

Question 2:

Answer:

By second fundamental theorem of calculus, we obtain

Question 3:

Answer:

By second fundamental theorem of calculus, we obtain

Question 4:

Answer:

By second fundamental theorem of calculus, we obtain

Question 5:

Answer:

By second fundamental theorem of calculus, we obtain

Question 6:

Answer:

By second fundamental theorem of calculus, we obtain

Question 7:

Answer:

By second fundamental theorem of calculus, we obtain

Question 8:

Answer:

By second fundamental theorem of calculus, we obtain

Question 9:

Answer:

By second fundamental theorem of calculus, we obtain

Question 10:

Answer:

By second fundamental theorem of calculus, we obtain

Question 11:

Answer:

By second fundamental theorem of calculus, we obtain

Question 12:

Answer:

By second fundamental theorem of calculus, we obtain

Question 13:

Answer:

By second fundamental theorem of calculus, we obtain

Question 14:

Answer:

By second fundamental theorem of calculus, we obtain

Question 15:

Answer:

By 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 I1 in (1), we obtain

Question 17:

Answer:

By second fundamental theorem of calculus, we obtain

Question 18:

Answer:

By second fundamental theorem of calculus, we obtain

Question 19:

Answer:

By second fundamental theorem of calculus, we obtain

Question 20:

Answer:

By second fundamental theorem of calculus, we obtain

Question 21:

equals

A. 

B. 

C. 

D.

Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

Question 22:

equals

A. 

B. 

C. 

D. 

Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is C.

Page No 340:

Question 1:

Answer:

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

Question 2:

Answer:

Also, let 

Question 3:

Answer:

Also, let x = tanθ ⇒ dx = sec2θ dθ

When x = 0, θ = 0 and when = 1, 

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

Question 4:

Answer:

Let + 2 = t2 ⇒ dx = 2tdt

When x = 0,  and when = 2, = 2

Question 5:

Answer:

Let cos x = t ⇒ −sinx dx = dt

When x = 0, = 1 and when

Question 6:

Answer:

Let ⇒ dx = dt

Question 7:

Answer:

Let x + 1 = ⇒ dx = dt

When x = −1, = 0 and when x = 1, = 2

Question 8:

Answer:

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

Answer:

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

Answer:

Integrating by parts, we obtain

Hence, the correct answer is B.

Page No 347:

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:

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

Question 6:

Answer:

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:

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:

Answer:

Adding (1) and (2), we obtain

Question 13:

Answer:

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:

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 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:

Answer:

It is known that, 

Adding (1) and (2), we obtain

Question 18:

Answer:

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

D. 

Answer:

Adding (1) and (2), we obtain

Hence, the correct answer is C.

Page No 352:

Question 1:

Answer:

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:

Answer:

Question 3:

 [Hint: Put]

Answer:

Question 4:

Answer:

Question 5:

Answer:

On dividing, we obtain

Question 6:

Answer:

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:

Answer:

Let  a ⇒ 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= t ⇒ 4x3 dx = dt

Question 13:

Answer:

Let ex = t ⇒ ex dx = dt

Question 14:

Answer:

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:

Answer:

= cos3 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 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

Page No 353:

Question 23:

Answer:

Question 24:

Answer:

Integrating by parts, we obtain

Question 25:

Answer:

Question 26:

Answer:

When = 0, = 0 and 

Question 27:

Answer:

When 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 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:

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 − x2 = t ⇒ −2x dx = dt

Hence, the given result is proved.

Question 40:

Evaluate as a limit of a sum.

Answer:

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.

Page No 354:

Question 43:

If 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.

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