#### 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 .

#### Question 4:

#### Answer:

The anti derivative of

is the function of *x *whose derivative is .

It is known that,

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.

#### Page No 304:

#### 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 I=∫xx+4 dxput x+4=t⇒dx=dtNow, I=∫t-4tdt=∫t-4t-1/2dt=23t3/2-42t1/2+C=23.t.t1/2-8t1/2+C=23x+4x+4-8x+4+C=23xx+4+83x+4-8x+4+C=23xx+4-163x+4+C=23x+4x-8+C

#### 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*

#### Page No 305:

#### 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*

⇒∫tan22x-3dx = ∫sec22x-3 – 1dx=∫sec2t- 1dt2= 12∫sec2t dt – ∫1dt= 12tant – t + C= 12tan2x-3 – 2x-3 + C

#### 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.

#### Page No 307:

#### 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,

sin A . sin B = 12cosA-B-cosA+B∴∫sin4x sin8x dx=∫12cos4x-8x-cos4x+8xdx=12∫cos-4x-cos12xdx=12∫cos4x-cos12xdx=12sin4x4-sin12x12+C

#### 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^{4}*x*

#### Answer:

From equation (1), we obtain

#### Question 17:

#### Answer:

#### Question 18:

#### Answer:

#### Question 19:

#### Answer:

1sinxcos3x=sin2x+cos2xsinxcos3x=sinxcos3x+1sinxcosx

⇒1sinxcos3x=tanxsec2x+1cos2xsinxcosxcos2x=tanxsec2x+sec2xtanx

#### Question 20:

#### Answer:

#### Question 21:

sin^{−1} (cos* x*)

#### Answer:

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 *x* + sec* x* + C

#### Answer:

Hence, the correct answer is A.

#### Question 24:

equals

**A.** âˆ’ 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*

#### 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*

#### Page No 316:

#### Question 10:

#### Answer:

#### Question 11:

19×2+6x+5

#### Answer:

∫19×2+6x+5dx=∫13x+12+22dx

Let (3x+1)=t

∴

3 dx=dt

⇒∫13x+12+22dx=13∫1t2+22dt

=13×2tan-1t2+C

=16tan-13x+12+C

#### 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^{−1}* x* + C

#### Answer:

Hence, the correct answer is B.

#### 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

#### 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 − *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

#### 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 (*x*^{3} +* x *+ 1) by *x*^{2} − 1, we obtain

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}^{ − 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

#### 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

#### 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} = *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

#### 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.

#### 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

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

#### 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

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

#### Question 8:

#### Answer:

Let

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

#### Question 9:

#### Answer:

Let

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

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

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

I=log x 2∫xdx-∫ddxlog x 2∫xdxdx=x22log x 2-∫2log x .1x.x22dx=x22log x 2-∫xlog x dx

Again integrating by parts, we obtain

I = x22logx 2-log x ∫x dx-∫ddxlog x ∫x dxdx=x22logx 2-x22log x -∫1x.x22dx=x22logx 2-x22log x +12∫x dx=x22logx 2-x22log x +x24+C

#### 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,

#### 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.

#### Page No 330:

#### 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.

#### Page No 334:

#### 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,

#### Page No 338:

#### 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 *x* and constant term, we obtain

A = 10 and B = −25

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.

#### Page No 340:

#### 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*

#### 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** d**θ*= *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.

#### Page No 347:

#### 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

x=π2, t=π∴

I=12∫0πlog sin tdt-π2log 2

⇒I=I2-π2log 2 [from 3]

⇒I2=-π2log 2

⇒I=-πlog 2

#### 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.

#### 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

#### 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} = *t* ⇒ *e*^{x} *dx* = *dt*

#### 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:

Let I=∫sin-1x-cos-1xsin-1x+cos-1xdx

It is known that, sin-1x+cos-1x=π2

⇒I=∫π2-cos-1x-cos-1xπ2dx

=2π∫π2-2cos-1xdx

=2π.π2∫1.dx-4π∫cos-1xdx

=x-4π∫cos-1xdx …(1)

Let I1=∫cos-1x dx

Also, let x=t⇒dx=2 t dt

⇒I1=2∫cos-1t.t dt

=2cos-1t.t22-∫-11-t2.t22dt

=t2cos-1t+∫t21-t2dt

=t2cos-1t-∫1-t2-11-t2dt

=t2cos-1t-∫1-t2dt+∫11-t2dt

=t2cos-1t-t21-t2-12sin-1t+sin-1t

=t2cos-1t-t21-t2+12sin-1t

From equation (1), we obtain

I=x-4πt2cos-1t-t21-t2+12sin-1t =x-4πxcos-1x-x21-x+12sin-1x

=x-4πxπ2-sin-1x-x-x22+12sin-1x

#### 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

#### Page No 353:

#### 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

Hence, the given result is proved.

#### 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:

Hence, the given result is proved.

#### 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.

#### Page No 354:

#### 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.