Class XII Chapter 7 – Integrals Maths

Exercise 7.1

Question 1: sin 2*x *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

Question 2:

Cos 3*x *Answer

The anti derivative of cos 3

Therefore, the anti derivative of .

Question 3:

*e*2*x*

Answer

The anti derivative of *e*^{2x}* *is
the function of *x *whose
derivative is *e*^{2x}. It is known
that,

Therefore, the anti derivative of .

Question 4:

Answer

The anti derivative of is the function of

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.

Exercise 7.2

Question 1:

Answer

Let = *t*

* *

* *

∴2*x 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

sin *x *⋅ sin (cos *x*)

Let cos *x *=
*t*

* *

* *

∴ −sin *x dx = dt*

* *

* *

* *

* *

* *

Question 5:

Answer

Let

∴ 2*adx = dt*

* *

* *

* *

Question 6:

Answer

Let *ax + b = t*

* *

* *

⇒ *adx = dt*

* *

* *

* *

* *

* *

* *

* *

Question 7:

Answer Let

∴ *dx = dt*

* *

* *

* *

* *

Question 8:

Answer

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

* *

* *

∴ 4*xdx = dt*

* *

* *

* *

Question 9:

Answer Let

∴ (2*x *+ 1)*dx
= dt*

* *

* *

* *

* *

* *

Question 10:

Answer

Let

∴

Question 11:

Answer Let

∴ *dx = dt*

* *

* *

* *

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

∴ 2*dx = dt*

* *

* *

* *

* *

* *

Question 17:

Answer Let

∴ 2*xdx = dt*

* *

* *

* *

* *

* *

* *

Question 18:

Answer Let

∴

Question 19:

Answer

Dividing numerator and denominator by *e ^{x}*, we obtain

Let

∴

Question 20:

Answer Let

∴

Question 21:

Answer

Let 2*x *− 3 = *t*

* *

* *

∴ 2*dx = dt*

* *

* *

* *

Question 22:

Answer

Let 7 −
4*x *= *t*

* *

* *

∴ −4*dx = dt*

* *

* *

* *

* *

* *

Question 23:

Answer Let

∴

Question 24:

Answer

Let

∴

Question 25:

Answer

Let

∴

Question 26:

Answer Let

∴

Question 27:

Answer

Let sin 2*x *= *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 *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

∴

Question 37:

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

* *

∴ 4*x*^{3}
*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.

Exercise 7.3

Question 1:

Answer

Question 2:

Answer

It is known that,

Question 3:

cos 2*x *cos
4*x *cos 6*x
*Answer

It is known that,

Question 4: sin^{3} (2*x *+ 1) Answer

Let

Question 5: sin^{3}
*x *cos^{3} *x *Answer

Question 6:

sin
*x *sin 2*x *sin 3*x
*Answer

It is known that,

Question 7: sin
4*x *sin 8*x *Answer

It is known that,

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:

tan

From equation (1), we obtain

Question 17:

Answer

Question 18:

Answer

Question 19:

Answer

Question 20:

Answer

Question 21: sin

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

Hence, the correct Answer is A.

Question 24:

equals

A. − cot (*ex ^{x}*) + C

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

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

D.

cot (

Let *ex ^{x} *=

* *

Hence, the correct Answer is B.

Exercise 7.4

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:

Answer

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

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

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

Hence, the correct Answer is B.

Question 25:

equals

A.

B.

C.

D.

Answer

Hence, the correct Answer is B.

Exercise 7.5

Question 1:

Answer

Let

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

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

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 −

Let

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

*A *= 2 and
*B *= 3

Substituting in 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

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

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 (

Let

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

On solving these equations, we obtain

Question 16:

[Hint: multiply
numerator and denominator by *x ^{n} *

Answer

Multiplying numerator
and denominator by *x ^{n} *

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 *⇒ 2*x dx *=
*dt*

* *

* *

* *

* *

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

Question 20:

Answer

Multiplying 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} *=

Answer

Let
*e ^{x} *=

* *

* *

* *

* *

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.

Exercise 7.6

Question 1:

*x *sin *x*

Answer

Let *I *=

Taking

Question 2:

Answer

Let *I *=

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

Question 5:

*x *log
2*x *Answer

Let

Taking log 2

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

Question 8:

Answer Let

Taking as first function and

Question 9:

Answer

Let

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

Page 106 of 216

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

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 1 as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

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

⇒

⇒

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 ⇒

= 2*θ*

* *

* *

⇒

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.

Exercise 7.7

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.

Exercise 7.8

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,

Exercise 7.9

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

Substituting the value of *I*_{1} 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.

Exercise 7.10

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 *= tan*θ *⇒ *dx *= sec^{2}*θ *d*θ*

* *

* *

When *x *= 0, *θ *= 0 and when *x *= 1,

Taking*θ*as first function and sec^{2}*θ *as
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*

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

* *

Page 153 of 216

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 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.

Exercise 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

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^{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

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

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.

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} *=

* *

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

From equation (1), we obtain

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 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

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

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

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 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.