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#### 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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