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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

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 .

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 .

#### Answer:

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

It is known that,

Therefore, the anti derivative of .

#### Answer:

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

It is known that,

Therefore, the anti derivative of is .

#### Answer:

On dividing, we obtain

#### 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.

Let = t

∴2x dx = dt

Let log |x| = t

#### Answer:

Let 1 + log x = t

#### Question 4:

sin x ⋅ sin (cos x)

#### Answer:

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt

Let

∴ 2adx = dt

Let ax + b = t

adx = dt

Let

dx = dt

Let 1 + 2x2 = t

∴ 4xdx = dt

#### Answer:

Let

∴ (2x + 1)dx = dt

Let

#### Answer:

Let

Let

∴ 9x2 dx = dt

Let log x = t

Let

∴ −8x dx = dt

Let

∴ 2dx = dt

Let

∴ 2xdx = dt

Let

#### Answer:

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

cot x log sin x

#### Answer:

Let log sin x = t

#### Answer:

Let 1 + cos x = t

∴ −sin x dx = dt

#### Answer:

Let 1 + cos x = t

∴ −sin x dx = dt

#### Answer:

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

#### Answer:

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

#### Answer:

Let 1 + log x = t

Let

#### Answer:

Let x4 = t

∴ 4x3 dx = dt

Let

From (1), we obtain

equals

#### Answer:

Let

Hence, the correct answer is D.

equals

A.

B.

C.

D.

#### Answer:

Hence, the correct answer is B.

#### Answer:

It is known that,

#### Question 3:

cos 2x cos 4x cos 6x

#### Answer:

It is known that,

sin3 (2x + 1)

Let

sin3 x cos3 x

#### Question 6:

sin x sin 2x sin 3x

#### Answer:

It is known that,

sin 4x sin 8x

#### Answer:

It is known that,

sin4 x

cos4 2x

tan4x

#### Answer:

From equation (1), we obtain

#### Answer:

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

sin−1 (cos x)

#### Answer:

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

#### Answer:

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

#### Answer:

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

#### Answer:

From (1), we obtain

Let x3 = t

⇒ 3x2 dx = dt

Let tan x = t

∴ sec2x dx = dt

#### Question 11:

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

#### Answer:

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

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

#### 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

#### Answer:

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

From (1), we obtain

From equation (2), we obtain

#### Answer:

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

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

#### 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

#### Answer:

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

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

#### Answer:

Let x2 + 2x +3 = t

⇒ (2x + 2) dx =dt

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

#### Answer:

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

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

#### 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.

equals

A.

B.

C.

D.

#### Answer:

Hence, the correct answer is B.

#### 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

#### Answer:

Let

Equating the coefficients of x and constant term, we obtain

A + B = 0

−3A + 3B = 1

On solving, we obtain

#### Answer:

Let

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

A = 1, B = −5, and C = 4

#### Answer:

Let

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

#### Answer:

Let

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

A = −2 and B = 4

#### 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

A = 2 and B = 3

Substituting in equation (1), we obtain

#### Answer:

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

#### 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

#### 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

#### Answer:

Let

Equating the coefficients of x2 and x, we obtain

#### Answer:

Let

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

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

#### Answer:

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

#### Answer:

Equating the coefficient of x and constant term, we obtain

A = 3

2A + B = −1 ⇒ B = −7

#### Answer:

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]

#### Answer:

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]

#### Answer:

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

A = 1 and B = −1

#### Answer:

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

#### Answer:

Let x2 = t ⇒ 2x dx = dt

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

#### 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 = t ex dx = dt

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

A = −1 and B = 1

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.

A.

B.

C.

D.

#### Answer:

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.

x sin x

#### Answer:

Let I =

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

#### Answer:

Let I =

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

#### Answer:

Let

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

Again integrating by parts, we obtain

x logx

#### Answer:

Let

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

x log 2x

#### Answer:

Let

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

x2 log x

#### Answer:

Let

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

#### Answer:

Let

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

#### Answer:

Let

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

#### Answer:

Let

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

#### Answer:

Let

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

#### Answer:

Let

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

#### Answer:

Let

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

#### Answer:

Let

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

#### Answer:

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

Again integrating by parts, we obtain

#### Answer:

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

#### Answer:

Let

Let

It is known that,

#### Answer:

Let

Let

It is known that,

#### Answer:

Let

It is known that,

From equation (1), we obtain

#### Answer:

Also, let

It is known that,

#### Answer:

Let

It is known that,

#### Answer:

Let

Integrating by parts, we obtain

Again integrating by parts, we obtain

#### Answer:

Let

= 2θ

Integrating by parts, we obtain

equals

#### Answer:

Let

Also, let

Hence, the correct answer is A.

equals

#### Answer:

Let

Also, let

It is known that,

Hence, the correct answer is B.

is equal to

A.

B.

C.

D.

#### Answer:

Hence, the correct answer is A.

is equal to

A.

B.

C.

D.

#### Answer:

Hence, the correct answer is D.

#### Answer:

It is known that,

#### Answer:

It is known that,

#### Answer:

It is known that,

#### Answer:

It is known that,

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

#### Answer:

It is known that,

#### Answer:

It is known that,

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### 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

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Answer:

By second fundamental theorem of calculus, we obtain

equals

A.

B.

C.

D.

#### Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

equals

A.

B.

C.

D.

#### Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is C.

#### Answer:

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

Also, let

#### Answer:

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

#### Answer:

Let x + 2 = t2dx = 2tdt

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

#### Answer:

Let cos x = t ⇒ −sinx dx = dt

When x = 0, t = 1 and when

Let dx = dt

#### Answer:

Let x + 1 = t dx = dt

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

#### 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 cot θ = t  ⇒ −cosecθ 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

#### Answer:

Integrating by parts, we obtain

Hence, the correct answer is B.

#### Answer:

Adding (1) and (2), we obtain

#### Answer:

Adding (1) and (2), we obtain

#### Answer:

Adding (1) and (2), we obtain

#### Answer:

Adding (1) and (2), we obtain

#### Answer:

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

#### Answer:

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

#### Answer:

Adding (1) and (2), we obtain

#### Answer:

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

#### Answer:

Adding (1) and (2), we obtain

#### Answer:

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

#### Answer:

It is known that,

#### Answer:

Adding (1) and (2), we obtain

#### 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, t = 0
and when

#### Answer:

It is known that,

Adding (1) and (2), we obtain

#### 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

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.

The value of is

A. 2

B.

C. 0

D.

#### Answer:

Adding (1) and (2), we obtain

Hence, the correct answer is C.

#### Answer:

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]

#### Answer:

On dividing, we obtain

#### Answer:

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

#### Answer:

Let x a = t dx = dt

#### Answer:

Let sin x = t ⇒ cos x dx = dt

#### Answer:

Let x4 = t ⇒ 4x3 dx = dt

#### Answer:

Let ex = tex dx = dt

#### Answer:

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

#### Answer:

= cos3 x × sin x

Let cos x = t ⇒ −sin x dx = dt

#### Answer:

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

From equation (1), we obtain

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

#### Answer:

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

#### Answer:

Integrating by parts, we obtain

#### Answer:

When x = 0, t = 0 and

When and when

#### Answer:

When and when

As , therefore, is an even function.

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

#### Answer:

From equation (1), we obtain

#### Answer:

Adding (1) and (2), we obtain

#### Answer:

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

#### Answer:

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.

#### Answer:

Integrating by parts, we obtain

Hence, the given result is proved.

#### 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.

#### Answer:

Hence, the given result is proved.

#### Answer:

Hence, the given result is proved.

#### 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,

is equal to

A.

B.

C.

D.

#### Answer:

Hence, the correct answer is A.

is equal to

A.

B.

C.

D.

#### Answer:

Hence, the correct answer is B.

#### Question 43:

If then is equal to

A.

B.

C.

D.

#### Answer:

Hence, the correct answer is D.

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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