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 Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

#### Page No 299:

#### 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

#### Page No 299:

#### Question 2:

Cos
3*x*

#### Answer:

The anti derivative of
cos 3*x* is a function of *x* whose derivative is cos 3*x*.

It is known that,

Therefore, the anti derivative of .

#### Page No 299:

#### Question 3:

*e*^{2}^{x}

#### Answer:

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

It is known that,

Therefore, the anti derivative of .

#### Page No 299:

#### Question 4:

#### Answer:

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

It is known that,

Therefore, the anti derivative of .

#### Page No 299:

#### 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 .

#### Page No 299:

#### Question 6:

#### Answer:

#### Page No 299:

#### Question 7:

#### Answer:

#### Page No 299:

#### Question 8:

#### Answer:

#### Page No 299:

#### Question 9:

#### Answer:

#### Page No 299:

#### Question 10:

#### Answer:

#### Page No 299:

#### Question 11:

#### Answer:

#### Page No 299:

#### Question 12:

#### Answer:

#### Page No 299:

#### Question 13:

#### Answer:

On dividing, we obtain

#### Page No 299:

#### Question 14:

#### Answer:

#### Page No 299:

#### Question 15:

#### Answer:

#### Page No 299:

#### Question 16:

#### Answer:

#### Page No 299:

#### Question 17:

#### Answer:

#### Page No 299:

#### Question 18:

#### Answer:

#### Page No 299:

#### Question 19:

#### Answer:

#### Page No 299:

#### Question 20:

#### Answer:

#### Page No 299:

#### Question 21:

The anti derivative of equals

**(A) ** **(B)
**

**(C) (D) **

#### Answer:

Hence, the correct answer is C.

#### Page No 299:

#### 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*

∴2*x dx* =
*dt*

#### Page No 304:

#### Question 2:

#### Answer:

Let log |*x*| = *t*

∴

#### Page No 304:

#### Question 3:

#### Answer:

Let 1 + log *x *=
*t*

∴

#### Page No 304:

#### Question 4:

sin *x* ⋅
sin (cos *x*)

#### Answer:

sin *x* ⋅
sin (cos *x*)

Let cos *x* = *t*

∴ −sin *x
dx = dt*

#### Page No 304:

#### Question 5:

#### Answer:

Let

∴ 2*adx = dt*

#### Page No 304:

#### Question 6:

#### Answer:

Let *ax + b = t*

⇒ *adx = dt*

#### Page No 304:

#### Question 7:

#### Answer:

Let

∴ *dx = dt*

#### Page No 304:

#### Question 8:

#### Answer:

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

∴ 4*xdx = dt*

#### Page No 304:

#### Question 9:

#### Answer:

Let

∴ (2*x* + 1)*dx = dt*

#### Page No 304:

#### Question 10:

#### Answer:

Let

∴

#### Page No 304:

#### Question 11:

#### Answer:

$\mathrm{Let}\mathrm{I}=\int \frac{x}{\left(x+4\right)}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{put}x+4=t\phantom{\rule{0ex}{0ex}}\Rightarrow dx=dt\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{I}=\int \frac{\left(t-4\right)}{\sqrt{t}}dt\phantom{\rule{0ex}{0ex}}=\int \left(\sqrt{t}-4{t}^{-1/2}\right)dt\phantom{\rule{0ex}{0ex}}=\frac{2}{3}{t}^{3/2}-4\left(2{t}^{1/2}\right)+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}.t.{t}^{1/2}-8{t}^{1/2}+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\left(x+4\right)\sqrt{x+4}-8\sqrt{x+4}+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}x\sqrt{x+4}+\frac{8}{3}\sqrt{x+4}-8\sqrt{x+4}+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}x\sqrt{x+4}-\frac{16}{3}\sqrt{x+4}+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\left(\sqrt{x+4}\right)\left(x-8\right)+\mathrm{C}$

#### Page No 304:

#### Question 12:

#### Answer:

Let

∴

#### Page No 304:

#### Question 13:

#### Answer:

Let

∴ 9*x*^{2}
*dx = dt*

#### Page No 304:

#### Question 14:

#### Answer:

Let log *x = t*

∴

#### Page No 304:

#### Question 15:

#### Answer:

Let

∴ −8*x
dx* = *dt*

#### Page No 304:

#### Question 16:

#### Answer:

Let

∴ 2*dx = dt*

#### Page No 304:

#### Question 17:

#### Answer:

Let

∴ 2*xdx = dt*

#### Page No 305:

#### Question 18:

#### Answer:

Let **
**

∴

#### Page No 305:

#### Question 19:

#### Answer:

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

Let

∴

#### Page No 305:

#### Question 20:

#### Answer:

Let

∴

#### Page No 305:

#### Question 21:

#### Answer:

Let 2*x* − 3 = *t*

∴ 2*dx** = dt*

$\Rightarrow \int {\mathrm{tan}}^{2}\left(2x-3\right)dx\phantom{\rule{0ex}{0ex}}=\int \left[{\mathrm{sec}}^{2}\left(2x-3\right)-1\right]dx\phantom{\rule{0ex}{0ex}}=\int \left[{\mathrm{sec}}^{2}t-1\right]\frac{dt}{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\int {\mathrm{sec}}^{2}t\mathit{}dt-\int 1dt\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{tan}t-t+\mathrm{C}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{tan}\left(2x-3\right)-\left(2x-3\right)+\mathrm{C}\right]$

#### Page No 305:

#### Question 22:

#### Answer:

Let 7 − 4*x*
= *t*

∴ −4*dx
= dt*

#### Page No 305:

#### Question 23:

#### Answer:

Let

∴

#### Page No 305:

#### Question 24:

#### Answer:

Let

∴

#### Page No 305:

#### Question 25:

#### Answer:

Let

∴

#### Page No 305:

#### Question 26:

#### Answer:

Let

∴

#### Page No 305:

#### Question 27:

#### Answer:

Let sin 2*x* = *t*

∴

#### Page No 305:

#### Question 28:

#### Answer:

Let

∴ cos *x dx*
= *dt*

#### Page No 305:

#### Question 29:

cot *x* log sin *x*

#### Answer:

Let log sin *x* =
*t*

#### Page No 305:

#### Question 30:

#### Answer:

Let 1 + cos *x = t*

∴ −sin *x
dx* = *dt*

#### Page No 305:

#### Question 31:

#### Answer:

Let 1 + cos *x* =
*t*

∴ −sin *x*
*dx = dt*

#### Page No 305:

#### Question 32:

#### Answer:

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

#### Page No 305:

#### Question 33:

#### Answer:

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

#### Page No 305:

#### Question 34:

#### Answer:

#### Page No 305:

#### Question 35:

#### Answer:

Let 1 + log *x* =
*t*

∴

#### Page No 305:

#### Question 36:

#### Answer:

Let**
**

∴

#### Page No 305:

#### Question 37:

#### Answer:

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

∴ 4*x*^{3}*
dx = dt*

Let

∴

From (1), we obtain

#### Page No 305:

#### Question 38:

equals

#### Answer:

Let

∴

Hence, the correct answer is D.

#### Page No 305:

#### Question 39:

equals

**A.**

**B.**

**C.**

**D. **

#### Answer:

Hence, the correct answer is B.

#### Page No 307:

#### Question 1:

#### Answer:

#### Page No 307:

#### Question 2:

#### Answer:

It is known that,

#### Page No 307:

#### Question 3:

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

#### Answer:

It is known that,

#### Page No 307:

#### Question 4:

sin^{3} (2*x* + 1)

#### Answer:

Let

#### Page No 307:

#### Question 5:

sin^{3} *x* cos^{3} *x*

#### Answer:

#### Page No 307:

#### Question 6:

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

#### Answer:

It is known that,

#### Page No 307:

#### Question 7:

sin 4*x* sin 8*x*

#### Answer:

It is known that,

$\mathrm{sin}\mathrm{A}.\mathrm{sin}\mathrm{B}=\frac{1}{2}\left[\mathrm{cos}\left(\mathrm{A}-\mathrm{B}\right)-\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)\right]\phantom{\rule{0ex}{0ex}}\therefore \int \mathrm{sin}4x\mathrm{sin}8xdx\phantom{\rule{0ex}{0ex}}=\int \frac{1}{2}\left\{\mathrm{cos}\left(4x-8x\right)-\mathrm{cos}\left(4x+8x\right)\right\}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int \left\{\mathrm{cos}\left(-4x\right)-\mathrm{cos}\left(12x\right)\right\}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int \left\{\mathrm{cos}4x-\mathrm{cos}12x\right\}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\frac{\mathrm{sin}4x}{4}-\frac{\mathrm{sin}12x}{12}\right)+\mathrm{C}$

#### Page No 307:

#### Question 8:

#### Answer:

#### Page No 307:

#### Question 9:

#### Answer:

#### Page No 307:

#### Question 10:

sin^{4} *x*

#### Answer:

#### Page No 307:

#### Question 11:

cos^{4} 2*x*

#### Answer:

#### Page No 307:

#### Question 12:

#### Answer:

#### Page No 307:

#### Question 13:

#### Answer:

#### Page No 307:

#### Question 14:

#### Answer:

#### Page No 307:

#### Question 15:

#### Answer:

#### Page No 307:

#### Question 16:

tan^{4}*x*

#### Answer:

From equation (1), we obtain

#### Page No 307:

#### Question 17:

#### Answer:

#### Page No 307:

#### Question 18:

#### Answer:

#### Page No 307:

#### Question 19:

#### Answer:

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

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

#### Page No 307:

#### Question 20:

#### Answer:

#### Page No 307:

#### Question 21:

sin^{−1} (cos* x*)

#### Answer:

It is known that,

Substituting in equation (1), we obtain

#### Page No 307:

#### Question 22:

#### Answer:

#### Page No 307:

#### 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.

#### Page No 307:

#### Question 24:

equals

**A.** − 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*

Hence, the correct answer is B.

#### Page No 315:

#### Question 1:

#### Answer:

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

∴ 3*x*^{2} *dx* = *dt*

#### Page No 315:

#### Question 2:

#### Answer:

Let 2*x* = *t*

∴ 2*dx* = *dt*

#### Page No 315:

#### Question 3:

#### Answer:

Let 2 − *x *= *t*

⇒ −*dx* = *dt*

#### Page No 315:

#### Question 4:

#### Answer:

Let 5*x* =* t*

∴ 5*dx* = *dt*

#### Page No 315:

#### Question 5:

#### Answer:

#### Page No 315:

#### Question 6:

#### Answer:

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

∴ 3*x*^{2} *dx* = *dt*

#### Page No 315:

#### Question 7:

#### Answer:

From (1), we obtain

#### Page No 315:

#### Question 8:

#### Answer:

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

⇒ 3*x*^{2} *dx* = *dt*

#### Page No 315:

#### Question 9:

#### Answer:

Let tan *x* =* t*

∴ sec^{2}*x* *dx* = *dt*

#### Page No 316:

#### Question 10:

#### Answer:

#### Page No 316:

#### Question 11:

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

#### Answer:

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

$\mathrm{Let}(3x+1)=t$

∴ $3dx=dt$

$\Rightarrow \int \frac{1}{{\left(3x+1\right)}^{2}+{2}^{2}}dx=\frac{1}{3}\int \frac{1}{{t}^{2}+{2}^{2}}dt$

$=\frac{1}{3\times 2}{\mathrm{tan}}^{-1}\frac{t}{2}+C$

$=\frac{1}{6}{\mathrm{tan}}^{-1}\left(\frac{3x+1}{2}\right)+C$

#### Page No 316:

#### Question 12:

#### Answer:

#### Page No 316:

#### Question 13:

#### Answer:

#### Page No 316:

#### Question 14:

#### Answer:

#### Page No 316:

#### Question 15:

#### Answer:

#### Page No 316:

#### 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*

#### Page No 316:

#### 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

#### Page No 316:

#### 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

#### Page No 316:

#### 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

#### Page No 316:

#### 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

#### Page No 316:

#### Question 21:

#### Answer:

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

⇒ (2*x* + 2) *dx* =*dt*

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

#### Page No 316:

#### Question 22:

#### Answer:

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

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

#### Page No 316:

#### Question 23:

#### Answer:

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

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

#### Page No 316:

#### 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.

#### Page No 316:

#### 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* + *B *= 1

2*A* +* B *=
0

On solving, we obtain

*A* = −1 and
*B* = 2

#### Page No 322:

#### 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

#### Page No 322:

#### Question 3:

#### Answer:

Let

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

*A* = 1, *B*
= −5, and *C* = 4

#### Page No 322:

#### Question 4:

#### Answer:

Let

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

#### Page No 322:

#### Question 5:

#### Answer:

Let

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

*A* = −2 and
*B* = 4

#### Page No 322:

#### 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

Let

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

*A *= 2 and* B *=
3

Substituting in equation (1), we obtain

#### Page No 322:

#### 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

#### Page No 322:

#### 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

#### Page No 322:

#### 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

#### Page No 322:

#### Question 10:

#### Answer:

Let

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

#### Page No 322:

#### Question 11:

#### Answer:

Let

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

#### Page No 322:

#### 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

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

#### Page No 322:

#### 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

#### Page No 322:

#### Question 14:

#### Answer:

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

*A* = 3

2*A* + *B *=
−1 ⇒ *B* = −7

#### Page No 322:

#### Question 15:

#### Answer:

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

On solving these equations, we obtain

#### Page No 322:

#### Question 16:

[Hint:
multiply numerator and denominator by *x*^{n}^{
− 1} and put *x*^{n} = *t*]

#### Answer:

Multiplying numerator
and denominator by *x*^{n }^{− 1},
we obtain

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

*A* = 1 and *B*
= −1

#### Page No 322:

#### 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 *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

#### Page No 323:

#### Question 19:

#### Answer:

Let *x*^{2}
= *t* ⇒ 2*x* *dx*
= *dt*

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

#### Page No 323:

#### 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

#### Page No 323:

#### 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

#### Page No 323:

#### 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.

#### Page No 323:

#### 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.

#### 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

#### Page No 327:

#### Question 2:

#### Answer:

Let *I* =

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

#### Page No 327:

#### 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

#### Page No 327:

#### Question 4:

*x* log*x*

#### Answer:

Let

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

#### Page No 327:

#### 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

#### Page No 327:

#### 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

#### Page No 327:

#### Question 7:

#### Answer:

Let

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

#### Page No 327:

#### Question 8:

#### Answer:

Let

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

#### Page No 327:

#### Question 9:

#### Answer:

Let

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

#### Page No 327:

#### Question 10:

#### Answer:

Let

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

#### Page No 327:

#### Question 11:

#### Answer:

Let

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

#### Page No 327:

#### Question 12:

#### Answer:

Let

Taking *x* as
first function and sec^{2}*x* as second function and
integrating by parts, we obtain

#### Page No 327:

#### Question 13:

#### Answer:

Let

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

#### Page No 327:

#### Question 14:

#### Answer:

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

$I={\left(\mathrm{log}x\right)}^{2}\int xdx-\int \left[\left\{\frac{d}{dx}{\left(\mathrm{log}x\right)}^{2}\right\}\int xdx\right]dx\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}{\left(\mathrm{log}x\right)}^{2}-\left[\int 2\mathrm{log}x.\frac{1}{x}.\frac{{x}^{2}}{2}dx\right]\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}{\left(\mathrm{log}x\right)}^{2}-\int x\mathrm{log}xdx\phantom{\rule{0ex}{0ex}}$

Again integrating by parts, we obtain

$I\hspace{0.17em}=\frac{{x}^{2}}{2}{\left(\mathrm{log}x\right)}^{2}-\left[\mathrm{log}x\int xdx-\int \left\{\left(\frac{d}{dx}\mathrm{log}x\right)\int xdx\right\}dx\right]\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}{\left(\mathrm{log}x\right)}^{2}-\left[\frac{{x}^{2}}{2}\mathrm{log}x-\int \frac{1}{x}.\frac{{x}^{2}}{2}dx\right]\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}{\left(\mathrm{log}x\right)}^{2}-\frac{{x}^{2}}{2}\mathrm{log}x+\frac{1}{2}\int xdx\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{2}{\left(\mathrm{log}x\right)}^{2}-\frac{{x}^{2}}{2}\mathrm{log}x+\frac{{x}^{2}}{4}+C$

#### Page No 327:

#### 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

#### Page No 328:

#### Question 16:

#### Answer:

Let

Let

⇒

∴

It is known that,

#### Page No 328:

#### Question 17:

#### Answer:

Let

Let ⇒

It is known that,

#### Page No 328:

#### Question 18:

#### Answer:

Let**
**⇒

It is known that,

From equation (1), we obtain

#### Page No 328:

#### Question 19:

#### Answer:

Also, let ⇒

It is known that,

#### Page No 328:

#### Question 20:

#### Answer:

Let ⇒

It is known that,

#### Page No 328:

#### Question 21:

#### Answer:

Let

Integrating by parts, we obtain

Again integrating by parts, we obtain

#### Page No 328:

#### Question 22:

#### Answer:

Let ⇒

= 2*θ*

⇒

Integrating by parts, we obtain

#### Page No 328:

#### Question 23:

equals

#### Answer:

Let

Also, let ⇒

Hence, the correct answer is A.

#### Page No 328:

#### Question 24:

equals

#### Answer:

Let

Also, let ⇒

It is known that,

Hence, the correct answer is B.

#### Page No 330:

#### Question 1:

#### Answer:

#### Page No 330:

#### Question 2:

#### Answer:

#### Page No 330:

#### Question 3:

#### Answer:

#### Page No 330:

#### Question 4:

#### Answer:

#### Page No 330:

#### Question 5:

#### Answer:

#### Page No 330:

#### Question 6:

#### Answer:

#### Page No 330:

#### Question 7:

#### Answer:

#### Page No 330:

#### Question 8:

#### Answer:

#### Page No 330:

#### Question 9:

#### Answer:

#### Page No 330:

#### Question 10:

is equal to

**A.**

**B.**

**C.**

**D. **

#### Answer:

Hence, the correct answer is A.

#### Page No 330:

#### 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,

#### Page No 334:

#### Question 2:

#### Answer:

It is known that,

#### Page No 334:

#### Question 3:

#### Answer:

It is known that,

#### Page No 334:

#### Question 4:

#### Answer:

It is known that,

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

#### Page No 334:

#### Question 5:

#### Answer:

It is known that,

#### Page No 334:

#### Question 6:

#### Answer:

It is known that,

#### Page No 338:

#### Question 1:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 2:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 3:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 4:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 5:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 6:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 7:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 8:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 9:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 10:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 11:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 12:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 13:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 14:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 15:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### 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

#### Page No 338:

#### Question 17:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 18:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 19:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 20:

#### Answer:

By second fundamental theorem of calculus, we obtain

#### Page No 338:

#### Question 21:

equals

**A.**

**B.**

**C.**

**D. **

#### Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

#### Page No 338:

#### 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

#### Page No 340:

#### Question 2:

#### Answer:

Also, let

#### Page No 340:

#### 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

#### Page No 340:

#### Question 4:

#### Answer:

Let *x *+ 2 = *t*^{2}
⇒ *dx *= 2*tdt*

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

#### Page No 340:

#### Question 5:

#### Answer:

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

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

#### Page No 340:

#### Question 6:

#### Answer:

Let
⇒
*dx* = *dt*

#### Page No 340:

#### Question 7:

#### Answer:

Let
*x* + 1 = *t *⇒ *dx*
= *dt*

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

#### Page No 340:

#### Question 8:

#### Answer:

Let 2*x* =* t*
⇒ 2*dx* = *dt*

When *x* = 1,* t*
= 2 and when *x* = 2, *t* = 4

#### Page No 340:

#### Question 9:

The value of the integral is

**A. **6

**B. **0

**C. **3

**D.** 4

#### Answer:

Let cot *θ* =* t *⇒ −cosec^{2 }*θ** **d**θ*= *dt*

Hence, the correct answer is A.

#### Page No 340:

#### 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.

#### Page No 347:

#### Question 1:

#### Answer:

Adding (1) and (2), we obtain

#### Page No 347:

#### Question 2:

#### Answer:

Adding (1) and (2), we obtain

#### Page No 347:

#### Question 3:

#### Answer:

Adding (1) and (2), we obtain

#### Page No 347:

#### Question 4:

#### Answer:

Adding (1) and (2), we obtain

#### Page No 347:

#### Question 5:

#### Answer:

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

#### Page No 347:

#### Question 6:

#### Answer:

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

#### Page No 347:

#### Question 7:

#### Answer:

#### Page No 347:

#### Question 8:

#### Answer:

#### Page No 347:

#### Question 9:

#### Answer:

#### Page No 347:

#### Question 10:

#### Answer:

Adding (1) and (2), we obtain

#### Page No 347:

#### 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

#### Page No 347:

#### Question 12:

#### Answer:

Adding (1) and (2), we obtain

#### Page No 347:

#### 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

#### Page No 347:

#### Question 14:

#### Answer:

It is known that,

#### Page No 347:

#### Question 15:

#### Answer:

Adding (1) and (2), we obtain

#### Page No 347:

#### 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 $x=\frac{\mathrm{\pi}}{2},t=\mathrm{\pi}$

∴ $I=\frac{1}{2}{\int}_{0}^{\pi}\mathrm{log}\mathrm{sin}tdt-\frac{\mathrm{\pi}}{2}\mathrm{log}2$

$\Rightarrow I=\frac{I}{2}-\frac{\mathrm{\pi}}{2}\mathrm{log}2[\mathrm{from}3]$

$\Rightarrow \frac{I}{2}=-\frac{\mathrm{\pi}}{2}\mathrm{log}2$

$\Rightarrow I=-\mathrm{\pi log}2$

#### Page No 347:

#### Question 17:

#### Answer:

It is known that,

Adding (1) and (2), we obtain

#### Page No 347:

#### Question 18:

#### Answer:

It can be seen that, (*x* − 1) ≤ 0 when 0 ≤ *x* ≤ 1 and (*x* − 1) ≥ 0 when 1 ≤ *x* ≤ 4

#### Page No 347:

#### Question 19:

Show that
if
*f* and *g* are defined as
and

#### Answer:

Adding (1) and (2), we obtain

#### Page No 347:

#### 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.

#### Page No 347:

#### 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 *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

#### Page No 352:

#### Question 2:

#### Answer:

#### Page No 352:

#### Question 3:

[Hint: Put]

#### Answer:

#### Page No 352:

#### Question 4:

#### Answer:

#### Page No 352:

#### Question 5:

#### Answer:

On dividing, we obtain

#### Page No 352:

#### 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

#### Page No 352:

#### Question 7:

#### Answer:

Let *x *−*
a *= *t *⇒ *dx*
= *dt*

#### Page No 352:

#### Question 8:

#### Answer:

#### Page No 352:

#### Question 9:

#### Answer:

Let sin *x* = *t*
⇒ cos *x dx* = *dt*

#### Page No 352:

#### Question 10:

#### Answer:

#### Page No 352:

#### Question 11:

#### Answer:

#### Page No 352:

#### Question 12:

#### Answer:

Let *x*^{4 }=*
t* ⇒ 4*x*^{3}
*dx* = *dt*

#### Page No 352:

#### Question 13:

#### Answer:

Let *e*^{x}
= *t* ⇒ *e*^{x}
*dx* = *dt*

#### Page No 352:

#### 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

#### Page No 352:

#### Question 15:

#### Answer:

= cos^{3} *x* × sin *x*

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

$\Rightarrow \int {\mathrm{cos}}^{3}x{e}^{\mathrm{log}\mathrm{sin}x}dx=\int {\mathrm{cos}}^{3}x\mathrm{sin}xdx\phantom{\rule{0ex}{0ex}}=-\int {t}^{3}dt\phantom{\rule{0ex}{0ex}}=-\frac{{t}^{4}}{4}+c\phantom{\rule{0ex}{0ex}}=-\frac{{\mathrm{cos}}^{4}x}{4}+c$

#### Page No 352:

#### Question 16:

#### Answer:

#### Page No 352:

#### Question 17:

#### Answer:

#### Page No 352:

#### Question 18:

#### Answer:

$\mathrm{Let}\mathrm{I}=\int \frac{1}{\sqrt{{\mathrm{sin}}^{3}x.\mathrm{sin}\left(x+\alpha \right)}}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\int \frac{1}{\sqrt{{\mathrm{sin}}^{3}x\left(\mathrm{sin}x.\mathrm{cos}\alpha +\mathrm{cos}x.\mathrm{sin}\alpha \right)}}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\int \frac{1}{\sqrt{{\mathrm{sin}}^{4}x.\mathrm{cos}\alpha +{\mathrm{sin}}^{4}x\mathrm{cot}x.\mathrm{sin}\alpha}}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\int \frac{1}{{\mathrm{sin}}^{2}x\sqrt{\mathrm{cos}\alpha +\mathrm{sin}\alpha .\mathrm{cot}x}}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\int \frac{{\mathrm{cosec}}^{2}x}{\sqrt{\mathrm{cos}\alpha +\mathrm{sin}\alpha \mathrm{cot}x}}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{put}\mathrm{cos}\alpha +\mathrm{sin}\alpha \mathrm{cot}x=t\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(0-\mathrm{sin}\alpha {\mathrm{cosec}}^{2}x\right)dx=dt\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{cosec}}^{2}xdx=-\frac{1}{\mathrm{sin}\alpha}dt\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{so},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{I}=-\frac{1}{\mathrm{sin}\alpha}\int \frac{dt}{\sqrt{t}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-\frac{1}{\mathrm{sin}\alpha}\times 2\sqrt{t}+\mathrm{C}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-\frac{2}{\mathrm{sin}\alpha}\sqrt{\mathrm{cos}\alpha +\mathrm{sin}\alpha \mathrm{cot}x}+\mathrm{C}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-\frac{2}{\mathrm{sin}\alpha}\sqrt{\mathrm{cos}\alpha +\mathrm{sin}\alpha .\frac{\mathrm{cos}x}{\mathrm{sin}x}}+\mathrm{C}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-\frac{2}{\mathrm{sin}\alpha}\sqrt{\frac{\mathrm{sin}x.\mathrm{cos}\alpha +\mathrm{cos}x.\mathrm{sin}\alpha}{\mathrm{sin}x}}+\mathrm{C}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-\frac{2}{\mathrm{sin}\alpha}\sqrt{\frac{\mathrm{sin}\left(x+\alpha \right)}{\mathrm{sin}x}}+\mathrm{C}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{so},\phantom{\rule{0ex}{0ex}}\int \frac{1}{\sqrt{{\mathrm{sin}}^{3}x.\mathrm{sin}\left(x\mathit{}+\alpha \right)}}dx=-\frac{2}{\mathrm{sin}\alpha}\sqrt{\frac{\mathrm{sin}\left(x+\alpha \right)}{\mathrm{sin}x}}+\mathrm{C}\phantom{\rule{0ex}{0ex}}$

#### Page No 352:

#### Question 19:

#### Answer:

$\mathrm{Let}I=\int \frac{{\mathrm{sin}}^{-1}\sqrt{x}-{\mathrm{cos}}^{-1}\sqrt{x}}{{\mathrm{sin}}^{-1}\sqrt{x}+{\mathrm{cos}}^{-1}\sqrt{x}}dx$

$\mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that},{\mathrm{sin}}^{-1}\sqrt{x}+{\mathrm{cos}}^{-1}\sqrt{x}=\frac{\mathrm{\pi}}{2}$

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

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

$=\frac{2}{\mathrm{\pi}}.\frac{\mathrm{\pi}}{2}\int 1.dx-\frac{4}{\mathrm{\pi}}\int {\mathrm{cos}}^{-1}\sqrt{x}dx$

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

$\mathrm{Let}{I}_{1}=\int {\mathrm{cos}}^{-1}\sqrt{x}dx\phantom{\rule{0ex}{0ex}}$

$\mathrm{Also},\mathrm{let}\sqrt{x}=t\Rightarrow dx=2tdt$

$\Rightarrow {I}_{1}=2\int {\mathrm{cos}}^{-1}t.tdt$

$=2\left[{\mathrm{cos}}^{-1}t.\frac{{t}^{2}}{2}-\int \frac{-1}{\sqrt{1-{t}^{2}}}.\frac{{t}^{2}}{2}dt\right]$

$={t}^{2}{\mathrm{cos}}^{-1}t+\int \frac{{t}^{2}}{\sqrt{1-{t}^{2}}}dt\phantom{\rule{0ex}{0ex}}$

$={t}^{2}{\mathrm{cos}}^{-1}t-\int \frac{1-{t}^{2}-1}{\sqrt{1-{t}^{2}}}dt$

$={t}^{2}{\mathrm{cos}}^{-1}t-\int \sqrt{1-{t}^{2}}dt+\int \frac{1}{\sqrt{1-{t}^{2}}}dt$

$={t}^{2}{\mathrm{cos}}^{-1}t-\frac{t}{2}\sqrt{1-{t}^{2}}-\frac{1}{2}{\mathrm{sin}}^{-1}t+{\mathrm{sin}}^{-1}t$

$={t}^{2}{\mathrm{cos}}^{-1}t-\frac{t}{2}\sqrt{1-{t}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}t$

From equation (1), we obtain

$I=x-\frac{4}{\mathrm{\pi}}\left[{t}^{2}{\mathrm{cos}}^{-1}t-\frac{t}{2}\sqrt{1-{t}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}t\right]\phantom{\rule{0ex}{0ex}}=x-\frac{4}{\mathrm{\pi}}\left[x{\mathrm{cos}}^{-1}\sqrt{x}-\frac{\sqrt{x}}{2}\sqrt{1-x}+\frac{1}{2}{\mathrm{sin}}^{-1}\sqrt{x}\right]$

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

#### Page No 352:

#### Question 20:

#### Answer:

#### Page No 352:

#### Question 21:

#### Answer:

#### Page No 352:

#### 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

#### Page No 353:

#### Question 23:

#### Answer:

#### Page No 353:

#### Question 24:

#### Answer:

Integrating by parts, we obtain

#### Page No 353:

#### Question 25:

#### Answer:

#### Page No 353:

#### Question 26:

#### Answer:

When
*x *=
0, *t *=
0 and

#### Page No 353:

#### Question 27:

#### Answer:

When and when

#### Page No 353:

#### Question 28:

#### Answer:

When and when

As , therefore, is an even function.

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

#### Page No 353:

#### Question 29:

#### Answer:

#### Page No 353:

#### Question 30:

#### Answer:

#### Page No 353:

#### Question 31:

#### Answer:

From equation (1), we obtain

#### Page No 353:

#### Question 32:

#### Answer:

Adding (1) and (2), we obtain

#### Page No 353:

#### Question 33:

#### Answer:

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

#### Page No 353:

#### 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.

#### Page No 353:

#### Question 35:

#### Answer:

Integrating by parts, we obtain

Hence, the given result is proved.

#### Page No 353:

#### 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.

#### Page No 353:

#### Question 37:

#### Answer:

Hence, the given result is proved.

#### Page No 353:

#### Question 38:

#### Answer:

Hence, the given result is proved.

#### Page No 353:

#### Question 39:

#### Answer:

Integrating by parts, we obtain

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

Hence, the given result is proved.

#### Page No 353:

#### Question 40:

Evaluate as a limit of a sum.

#### Answer:

It is known that,

#### Page No 353:

#### Question 41:

is equal to

**A.**

**B.**

**C.**

**D. **

#### Answer:

Hence, the correct answer is A.

#### Page No 353:

#### 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.

#### Page No 354:

#### 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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