NCERT Solutions for Class 12 Science Math Chapter 1 Integrals are provided here with simple step-by-step explanations. These solutions for Integrals are extremely popular among Class 12 Science students for Math Integrals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

#### Question 1:

sin 2x

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

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

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

It is known that,

Therefore, the anti derivative of .

#### Question 4:

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

It is known that,

Therefore, the anti derivative of .

#### Question 5:

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

It is known that,

Therefore, the anti derivative of is .

#### Question 13:

On dividing, we obtain

#### Question 21:

The anti derivative 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,

∴Anti derivative of

Also,

Hence, the correct answer is A.

Let = t

∴2x dx = dt

Let log |x| = t

#### Question 3:

Let 1 + log x = t

#### Question 4:

sin x ⋅ sin (cos x)

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt

Let

Let ax + b = t

Let

dx = dt

Let 1 + 2x2 = t

∴ 4xdx = dt

#### Question 9:

Let

∴ (2x + 1)dx = dt

Let

#### Question 11:

Let

Let

∴ 9x2 dx = dt

Let log x = t

Let

∴ −8x dx = dt

Let

∴ 2dx = dt

Let

∴ 2xdx = dt

Let

#### Question 19:

Dividing numerator and denominator by ex, we obtain

Let

Let

Let 2x − 3 = t

∴ 2dx = dt

Let 7 − 4x = t

∴ −4dx = dt

Let

Let

Let

Let

Let sin 2x = t

Let

∴ 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 35:

Let 1 + log x = t

Let

#### Question 37:

Let x4 = t

∴ 4x3 dx = dt

Let

From (1), we obtain

#### Question 38:

equals

Let

Hence, the correct answer is D.

#### Question 39:

equals

A.

B.

C.

D.

Hence, the correct answer is B.

#### Question 2:

It is known that,

#### Question 3:

cos 2x cos 4x cos 6x

It is known that,

sin3 (2x + 1)

Let

sin3 x cos3 x

#### Question 6:

sin x sin 2x sin 3x

It is known that,

#### Question 7:

sin 4x sin 8x

It is known that,

sin4 x

cos4 2x

#### Question 16:

tan4x

From equation (1), we obtain

#### Question 19:

$\frac{1}{\mathrm{sin}x{\mathrm{cos}}^{3}x}=\frac{{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x}{\mathrm{sin}x{\mathrm{cos}}^{3}x}=\frac{\mathrm{sin}x}{{\mathrm{cos}}^{3}x}+\frac{1}{\mathrm{sin}x\mathrm{cos}x}$
$⇒\frac{1}{\mathrm{sin}x{\mathrm{cos}}^{3}x}=\mathrm{tan}x{\mathrm{sec}}^{2}x+\frac{\frac{1}{{\mathrm{cos}}^{2}x}}{\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{cos}}^{2}x}}=\mathrm{tan}x{\mathrm{sec}}^{2}x+\frac{{\mathrm{sec}}^{2}x}{\mathrm{tan}x}$

#### Question 21:

sin−1 (cos x)

It is known that,

Substituting in equation (1), we obtain

#### 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. − cot (exx) + C

B. tan (xex) + C

C. tan (ex) + C

D. cot (ex) + C

Let exx = t

Hence, the correct answer is B.

Let x3 = t

∴ 3x2 dx = dt

Let 2x = t

∴ 2dx = dt

Let 2 − x = t

⇒ −dx = dt

Let 5x = t

∴ 5dx = dt

Let x3 = t

∴ 3x2 dx = dt

#### Question 7:

From (1), we obtain

Let x3 = t

⇒ 3x2 dx = dt

Let tan x = t

∴ sec2x dx = dt

#### Question 11:

$\frac{1}{9{x}^{2}+6x+5}$

$\int \frac{1}{9{x}^{2}+6x+5}dx=\int \frac{1}{{\left(3x+1\right)}^{2}+{2}^{2}}dx$

$⇒\int \frac{1}{{\left(3x+1\right)}^{2}+{2}^{2}}dx=\frac{1}{3}\int \frac{1}{{t}^{2}+{2}^{2}}dt$
$=\frac{1}{3×2}{\mathrm{tan}}^{-1}\frac{t}{2}+C$
$=\frac{1}{6}{\mathrm{tan}}^{-1}\left(\frac{3x+1}{2}\right)+C$

#### 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

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

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.

#### Question 1:

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

#### Question 2:

Let

Equating the coefficients of x and constant term, we obtain

A + B = 0

−3A + 3B = 1

On solving, we obtain

#### Question 3:

Let

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

A = 1, B = −5, and C = 4

#### Question 4:

Let

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

#### Question 5:

Let

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

Let

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

A = 2 and B = 3

Substituting in equation (1), we obtain

#### Question 7:

Let

Equating the coefficients of x2, x, and constant term, we obtain

A + C = 0

A + B = 1

B + C = 0

On solving these equations, we obtain

From equation (1), we obtain

#### 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

Equating the coefficients of x2 and x, we obtain

#### Question 11:

Let

Substituting x = −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

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

#### Question 13:

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

AB = 0

BC = 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 + B = −1 ⇒ B = −7

#### Question 15:

Equating the coefficient of x3, x2, 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]

Multiplying numerator and denominator by xn − 1, we obtain

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

A = 1 and B = −1

#### Question 17:

[Hint: Put sin x = t]

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

A = 1 and B = −1

#### Question 18:

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

A + C = 0

B + D = 4

4A + 3C = 0

4B + 3D = 10

On solving these equations, we obtain

A = 0, B = −2, C = 0, and D = 6

#### Question 19:

Let x2 = t ⇒ 2x dx = dt

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

#### Question 20:

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]

Let ex = t 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.

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 x2, x, and constant term, we obtain

A + B = 0

C = 0

A = 1

On solving these equations, we obtain

A = 1, B = −1, and C = 0

Hence, the correct answer is A.

#### Question 1:

x sin x

Let I =

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

#### Question 2:

Let I =

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 logx

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:

x2 log 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

Again integrating by parts, we obtain

#### Question 15:

Let

Let I = I1 + I2 … (1)

Where, and

Taking log x as first function and x2 as second function and integrating by parts, we obtain

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

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

#### Question 16:

Let

Let

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

Let

Also, let

Hence, the correct answer is A.

#### Question 24:

equals

Let

Also, let

It is known that,

Hence, the correct answer is B.

#### 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.

#### Question 1:

It is known that,

#### Question 2:

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:

It is known that,

#### Question 6:

It is known that,

#### Question 1:

By second fundamental theorem of calculus, we obtain

#### Question 2:

By second fundamental theorem of calculus, we obtain

#### Question 3:

By second fundamental theorem of calculus, we obtain

#### Question 4:

By second fundamental theorem of calculus, we obtain

#### Question 5:

By second fundamental theorem of calculus, we obtain

#### Question 6:

By second fundamental theorem of calculus, we obtain

#### Question 7:

By second fundamental theorem of calculus, we obtain

#### Question 8:

By second fundamental theorem of calculus, we obtain

#### Question 9:

By second fundamental theorem of calculus, we obtain

#### Question 10:

By second fundamental theorem of calculus, we obtain

#### Question 11:

By second fundamental theorem of calculus, we obtain

#### Question 12:

By second fundamental theorem of calculus, we obtain

#### Question 13:

By second fundamental theorem of calculus, we obtain

#### Question 14:

By second fundamental theorem of calculus, we obtain

#### Question 15:

By 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 second fundamental theorem of calculus, we obtain

#### Question 18:

By second fundamental theorem of calculus, we obtain

#### Question 19:

By second fundamental theorem of calculus, we obtain

#### Question 20:

By second fundamental theorem of calculus, we obtain

#### Question 21:

equals

A.

B.

C.

D.

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

#### Question 22:

equals

A.

B.

C.

D.

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is C.

#### Question 1:

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

Also, let

#### Question 3:

Also, let x = tanθdx = 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

#### Question 4:

Let x + 2 = t2dx = 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

Let dx = dt

#### Question 7:

Let x + 1 = t dx = dt

When x = −1, t = 0 and when x = 1, t = 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 dθ= 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

Integrating by parts, we obtain

Hence, the correct answer is B.

#### 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:

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

#### Question 6:

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

#### Question 10:

Adding (1) and (2), we obtain

#### Question 11:

As sin2 (−x) = (sin (−x))2 = (−sin x)2 = sin2x, therefore, sin2x is 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:

As sin7 (−x) = (sin (−x))7 = (−sin x)7 = −sin7x, therefore, sin2x is 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, t = 0
and when

#### Question 17:

It is known that,

Adding (1) and (2), we obtain

#### Question 18:

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.

Adding (1) and (2), we obtain

Hence, the correct answer is C.

#### Question 1:

Equating the coefficients of x2, x, and constant term, we obtain

A + B C = 0

B + C = 0

A = 1

On solving these equations, we obtain

From equation (1), we obtain

[Hint: Put]

#### Question 5:

On dividing, we obtain

#### Question 6:

Equating the coefficients of x2, x, and constant term, we obtain

A + B = 0

B + C = 5

9A + C = 0

On solving these equations, we obtain

From equation (1), we obtain

#### Question 7:

Let x a = t dx = dt

#### Question 9:

Let sin x = t ⇒ cos x dx = dt

#### Question 12:

Let x4 = t ⇒ 4x3 dx = dt

#### Question 13:

Let ex = tex dx = dt

#### Question 14:

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

A + C = 0

B + D = 0

4A + C = 0

4B + D = 1

On solving these equations, we obtain

From equation (1), we obtain

#### Question 15:

= cos3 x × sin x

Let cos x = t ⇒ −sin x dx = dt

#### Question 19:

$⇒I=\int \frac{\left(\frac{\mathrm{\pi }}{2}-{\mathrm{cos}}^{-1}\sqrt{x}\right)-{\mathrm{cos}}^{-1}\sqrt{x}}{\frac{\mathrm{\pi }}{2}}dx$
$=\frac{2}{\mathrm{\pi }}\int \left(\frac{\mathrm{\pi }}{2}-2{\mathrm{cos}}^{-1}\sqrt{x}\right)dx$
$=\frac{2}{\mathrm{\pi }}.\frac{\mathrm{\pi }}{2}\int 1.dx-\frac{4}{\mathrm{\pi }}\int {\mathrm{cos}}^{-1}\sqrt{x}dx$

From equation (1), we obtain

$=x-\frac{4}{\mathrm{\pi }}\left[x\left(\frac{\mathrm{\pi }}{2}-{\mathrm{sin}}^{-1}\left(\sqrt{x}\right)\right)-\frac{\sqrt{x-{x}^{2}}}{2}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(\sqrt{x}\right)\right]$

#### Question 22:

Equating the coefficients of x2, x,and constant term, we obtain

A + C = 1

3A + B + 2C = 1

2A + 2B + C = 1

On solving these equations, we obtain

A = −2, B = 1, and C = 3

From equation (1), we obtain

#### Question 24:

Integrating by parts, we obtain

#### Question 26:

When x = 0, t = 0 and

When 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 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 x2, x, and constant term, we obtain

A + C = 0

A + B = 0

B = 1

On solving these equations, we obtain

A = −1, C = 1, and B = 1

Hence, the given result is proved.

#### 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.

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 then is equal to

A.

B.

C.

D.

Hence, the correct answer is D.

The value of is

A. 1

B. 0

C. − 1

D.