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RD Sharma Solutions Class 12 Mathematics Chapter 18 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 18 MCQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:30 PM IST

The class 12 RD Sharma chapter 18 exercise MCQ Indefinite Integrals course of action is particularly trusted and proposed by students and educators across the entire country The appropriate responses in the RD Sharma class 12th exercise MCQ are handpicked and made via prepared experts, making them exact and sensible enough for students In addition, in RD Sharma Solutions class 12 chapter 18 exercise MCQ, the experts offer response keys and some surprising tips in the book that the students presumably will not find elsewhere RD Sharma class 12 solutions MCQ Chapter 18 has around 43 questions

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  1. RD Sharma Class 12 Solutions Chapter 18 MCQ Indefinite Integrals - Other Exercise
  2. Indefinite Integrals Excercise: MCQ
  3. RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 18 MCQ Indefinite Integrals - Other Exercise

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Indefinite Integrals Excercise: MCQ

Indefinite Integrals Exercise Multiple Choice Questions Question 1

Answer:
14tan1x22+C
Given:
x4+x4dx
Hint:
Using dx1+x2
Explanation:
Let I=x4+x4dx
=xdx(2)2+(x2)2  [Put x2=t2x dx=dtxdx=dt2]
=1(2)2+t2×dt2=12dtt2+22=1212tan1t2+C..{1x2+a2dx=1atan1(xa)+c}=14tan1x22+C

Indefinite Integrals Exercise Multiple Choice Questions Question 2

Answer:
12log(tan(x2+π12))+C
Given:
1cosx+3sinxdx
Hint:
You must know about the cosecxdx
Explanation:
Let I=12(sinx82+cosx12)dx
=12dxsinxcos6π+cosxsinπ6 [cosπ6=32;sinπ6=12]
=12dxsin(x+π6) [sin(A+B)=sinAcosB+cosAsinB]
=12cosec(x+π6)dx [sinx=1cosθcx]
=12log(cosec(x+π6)+cot(x+π6))+C [cosec(x)dx=log(cosecx+cotx)+c]
Now,
cosecx+cotx=1sinx+cosxsinx=1+cosxsinx=2cos2x22sinx2cos2x [1+cos2θ=2cos2θ;sinθ=2sinθ2cosθ2]
=cotx2
 From Eq(i)
I=12log(cot(x+π62))+C=12log[cot(x2+π12)]1+C=12log(tan(x2+π12))+C

Indefinite Integrals Exercise Multiple Choice Questions Question 3

Answer:
12log(secx2+tanx2)+C
Given:
xsecx2dx
Hint:
Using secθdθ
Explanation:
Let,I=xsecx2dx [ Put x2=t2xdx=dtxdx=dt2]
=sectdt2=12sectdt=12log|sect+tant|+C {secxdx=log|secx+tanx|+c}
=12log|secx2+tanx2|+C

Indefinite Integrals Exercise Multiple Choice Questions Question 4

Answer:
A=23,B=53
Given:
15+4sinxdx=Atan1(Btanx2+43)+C
Hint
Using 11+t2dt
Explanation:
LetI=15+4sinxdx
=15+4.2sinx2cosx2dx=15.1+8sinx2cosx2dx [sin2θ=2sinθcosθ]
=15(sin2x2+cos2x22)+8sinx2cos2xdx [sin2θ+cos2θ=1]
=15sin2x2+5cos2x2+8sinx2cos2xdx
Divide num and den by cos2x2
=1cos2x25tan2x2+5+8tan2xdx=15sec2x2tan2x2+85tan2x+1dx [secx=1cosx]
=152t2+85t+1dt [ Put tanx2=tsec2x212dx=dtsec2x2dx=2dt]
=25dtt2+8t5+(45)2(45)2+1=25dt(t+45)2+(35)2=25×135tan1(5t+43)+C{1x2+a2dx=1atan1(xa)+c}
I=23tan1(53tanx2+43)+C [Eq(i)]
Acc to given,
I=Atan1(Btanx2+43)+C [Eq(ii)]

From Eq ( i ) and ( ii )

Atan1(Btanx2+43)+C=23tan1(53tanx2+43)+C
Comparing both sides, we get
A=23,B=53


Indefinite Integrals Exercise Multiple Choice Questions Question 5

Answer:
xsinx+C
Given:
xsinx(sinxx+cosxlogx)dx
Hint:
You must know about the derivation of xsinx
Explanation:
LetI=xsinx(sinxx+cosxlogx)
=tdtt  [ Put xsinx=t, Taking log of both sides 
=1dt sinxlogx=logt
=t+CDiff w r t t,
=xsinx+C(logxcosx+sinxx)dxdt=1t(logxcosx+sinxx)dx=dtt]


Indefinite Integrals Exercise Multiple Choice Questions Question 6

Answer:
tan1(logex)2+C
Given:
11+(logex)2dx w.r.t. (logex)
Hint:
Using11+t2dx
Explanation:
LetI=11+(logex)2dx(logx)
=11+(logex)2dxx [ddx(logx)=1xd(logx)=dxx]
=11+t2dt  [Put logex=t1xdx=dt]
=tan1t+C [11+x2dx=tan1x+C]
=tan1(logex)+C

Indefinite Integrals Exercise Multiple Choice Questions Question 7

Answer:
116
Given:
cos8x+1tan2xcot2xdx=acos8x+C
Hint
Using sinxdx
Explanation:
LetI=cos8x+1tan2xcot2xdx
Missing or unrecognized delimiter for \left
=2cos24xsin2xcos2xsin22xcos22xdx
=cos24xsin4xcos4xdx [cos2xsin2x=cos2x]
=cos4xsin4xdx
=12sin8xdx [sin2x=2sinxcosx]
=12sin8xdx [sin2x=2sinxcosx]
=12(cos8x8)+CI=cos8x16+C
According to given,
I=acos8x+Cacos8x+C=116cos8x+Ca=116

Indefinite Integrals Exercise Multiple Choice Questions Question 8

Answer:
12
Given:
sin8xcos8x12sin2xcos2xdx=asin2x+C
Hint:
Using identity (a2b2)&cosxdx
Explanation:
LetI=sin8xcos8x12sin2xcos2xdx
=(sin4x)2(cos4x)2(sin2x+cos2x)22sin2xcos2xdx [sin2x+cos2x=1]
=(sin4xcos4x)(sin4x+cos4x)(sin4x+cos4x)dx=(sin4xcos4x)dx=(cos4xsin4x)dx=(cos2xsin2x)(cos2x+sin2x)dx [a2b2=(a+b)(ab)]
=1cos2x(1)dx [sin2x+cos2x=1]
According to given : I=asin2x+C
asin2x+C=12sin2x+C
Comparing both sides, we get
a=12

Indefinite Integrals Exercise Multiple Choice Questions Question 9

Answer:
xex+C
Given:
(x1)exdx
Hint:
Using integration by parts & exdx
Explanation:
LetI=(x1)exdx
=xexdxexdx=xex1+(1)exdxexdx [u.vdx=uvdxdudxvdxdx]
=xex+exdxexdx [exdx=ex+C]
=xex+C

Indefinite Integrals Exercise Multiple Choice Questions Question 10

Answer:
1loge2
Given:
21xx2dx=K21x+C
Hint
Using axdx
Explanation:
Let=21xx2dx  [Put 1x=t1x2dx=dtdxx2=dt
=2tdt=2tloge2+2
=21xloge2+2 [axdx=axlogea]
According to given:I=K21x+C
K21x+C=1loge221x+C

Comparing both sides,

K=1loge2


Indefinite Integrals Exercise Multiple Choice Questions Question 11

Answer:
12[x+log|(sinx+cosx)|]+C
Given:
11+tanxdx
Hint:
You must know about the derivation of sin x and cos x and use dxx
Explanation:
Let I=11+tanxdx
=11+tanx
=11+sinxcosx [tanx=sinxcosx]
=cosxcosx+sinx
=2cosx2(cosx+sinx)
=(cosx+sinx)+(cosxsinx)2(cosx+sinx) [Special step]
=12[1+cosxsinxcosx+sinx]I2=11+tanxdx=12dx+12cosxsinxcosx+sinxdxI2=121dx+12dtt [Putcosx+sinx=t in I2]
[(sinx+cosx)dx=dt(cosxsinx)dx=dt]
=12x+12log|t|+C
=12[x+log|(cosx+sinx)|]+C

Indefinite Integrals Exercise Multiple Choice Questions Question 12

Answer:
None of these
Given:
|x|3dx
Hint:
Usingxndx
Explanation:
LetI=|x|3dx
Here two cases arise
I={x3,x<0;x3,x0 Case I, when x<0
I=4x3dx=x44+C [xndx=xn+1n+1+C]
 Case II, when x0
I=4x3dx=x44+C [xndx=xn+1n+1+C]
Combining these two, we get
I=±x44+C

Indefinite Integrals Exercise Multiple Choice Questions Question 13

Answer:
2sinx+C
Given:
cosxxdx
Hint
Using cosxdx
Explanation:
LetI=cosxxdx
=cost.2dt [ Put x=t12xdx=dtdxx=2dt]
=2costdt=2sint+C=2sinx+C [cosxdx=sinx+C]

Indefinite Integrals Exercise Multiple Choice Questions Question 14

Answer:
excotx+C
Given:
ex(1cotx+cot2x)dx
Hint:
Using integration by parts
Explanation:
Let I=ex(1cotx+cot2x)dx
=ex[(1+cot2x)cotx]dx=ex(cosec2xcotx)dx [1+cot2θ=cosec2θ]
=ex(cotxcosec2x)dx=excotxdx+excosec2xdx=[cotxex(cosec2x)exdx]+excosec2xdx.{uvdx=uvdxdudxvdxdx}=cotxexexcosec2xdx+excosec2xdx=cotxex+C [Using integration by parts ] 

Indefinite Integrals Exercise Multiple Choice Questions Question 15

Answer:
tan7x7+C
Given:
sin6xcos8xdx
Hint:
Using xndx
Explanation:
LetI=sin6xcos8xdx
=(sin6xcos6x×1cos2x)dx
=tan6xsec2xdx [tanx=sinxcosx;secx=1cosx]
=t6dt  Put tanx=tsec2xdx=dt]
=t77+C=tan7x7+C [xndx=xn+1n+1+C]

Indefinite Integrals Exercise Multiple Choice Questions Question 16

Answer:
16tan1(16tanx2)+C
Given:
(17+5cosx)dx
Hint:
You must know about the derivation of tan x and using 11+x2dx
Explanation:
Let I=17+5cosxdx
 Put cosx=1tan2x21+tan2x2
I=17+5(1tan2x21+tan2x2)dx=1+tan2x27(1+tan2x2)+55tan2x2dx=sec2x212+2tan2x2=12sec2x2tan2x2+6dx=122dtt2+(6)2  [ Put tanx2=tsec2x212dx=dtsec2x2dx=2dt]
=dtt2+(6)2=16tan1(t6)+C1x2+a2=1atan1xa+c=16tan1(tanx26)+C

Indefinite Integrals Exercise Multiple Choice Questions Question 17

Answer:
log|1cotx2|+C
Given:
11cosxsinxdx
Hint:
Using partial fraction and 1xdx
Explanation:
Let I=11cosxsinxdx
 Put cosx=1tan2x21+tanxx2;sinx=2tan2x1+tanx2I=11(1tan2x21+tan2x2)(2tanx21+tan2x2)dx
=1+tan2x21+tan2x21+tan22tanx2dx=sec2x22tan2x22tanx2dx=sec2x212tanx2(tan2x1)dx[1+tan2θ=sec2θ]
=dtt(t1)  [ Put tanx2=tsec2x212dx=dt]
Now,
1t(t1)=At+Bt1
Multiplying by t (t - 1)1=A(t1)+B(t)

Putting t = 1

1=A(01)+B(0)A=11t(t1)=1t+1t11t(t1)dt=1tdt+1t1dt
=log|t|+log|t1|+C=log|tanx2|+log|tanx21|+C=log|tanx21tan2xx|+C=log|1cotx2|+C[cotx=1tanx]


Indefinite Integrals Exercise Multiple Choice Questions Question 18

Answer:
exx+4+C
Given:
x+3(x+4)2exdx
Hint:
Using [u.vdx=uvdxdudxvdxdx
Explanation:
LetI=x+3(x+4)2exdx
=(x+3+11)(x+4)2exdx
=ex[x+4(x+4)21(x+4)2]dx
=ex[1x+41(x+4)2]dx=ex1x+4dxex1(x+4)2dx=1x+4ex1(x+4)2exdx1(x+4)2exdx..{u.vdx=uvdxdudxvdxdx}=exx+4+1(x+4)2exdx1(x+4)2exdx=exx+4+C

Indefinite Integrals Exercise Multiple Choice Questions Question 19

Answer:
123tan1(2cosx3)+C
Given:
sinx3+4cos2xdx
Hint:
You must know about the derivative of cos x and 11+x2dx
Explanation:
Let I=sinx3+4cos2xdx
=dt3+4t2 [ Put cosx=tsinxdx=dtsinxdx=dt]
=14dtt2+34=14dt(t)2+(32)2=14132tan1(t32)+c1x2+a2dx=tan1xa+c=123tan1(2t3)+C=123tan1(2cosx3)+C

Indefinite Integrals Exercise Multiple Choice Questions Question 20

Answer:
excotx2+C
Given:
ex(1sinx1cosx)dx
Hint:
You must know about the derivation of cot x and exdx
Explanation:
Let I=ex(1sinx1cosx)dx
=ex(sin2x2+cos2x22sinx2cos2x2sin2x2)dx [sin2θ+cos2θ=1;sin2θ=2sinθcosθ;1cosθ=2sin2θ2]
=12ex(sinx2cosx2sinx2)2dx=12ex(1cotx2)2dx=12ex((1+cot2x2)2cotx2)dx [cotx=cosxsinx]
=12ex(cosec2x22cotx2)dx [1+cot2θ=cose2θ]
=12excosec2x2dxcotx2exdx=cotx2exdx+cosec2x2ex12dx=[cotx2excosec2x212exdx]+12excosec2x2dx..[u.vdx=uvdxdudxvdxdx]=cotx2ex12excosec2x2dx+12excosec2x2dx=cotx2ex+C

Indefinite Integrals Exercise Multiple Choice Questions Question 21

Answer:
exex+ex+C
Given:
2(ex+ex)2dx
Hint:
You must know about the xndx
Explanation:
I=2dx(ex+ex)2I=2exex(ex+ex)2dx Put ex=texdx=dtI=2dtt(t+1t)2I=2dtt(t2+1)2t2=2tdt(t2+1)2 Put z=t2+1dz=2tdt
I=dzz2=z2dz=z2+12+1+C[xndx=xn+1n+1+C]=1z+C=1(t2+1)+C[z=t2+1]=1e2x+1+C[t=ex]=1ex(ex+1ex)+C=ex(ex+ex)+C
Hence I=exex+ex+C
Option (a) is correct

Indefinite Integrals Exercise Multiple Choice Questions Question 22

Answer:
tan(xex)+C
Given:
ex(1+x)cos2(xex)dx
Hint:
You must know about the sec2xdx
Explanation:
Let I=ex(1+x)cos2(xex)dx
=dtcos2t  [ Put xex=t(xex+ex(1))dx=dtex(x+1)dx=dt]
=sec2tdt [secθ=1cosθ]
=tant+C=tan(xex)+C
Hence,ex(1+x)dxcos2(xex)=tan(xex)+C

Indefinite Integrals Exercise Multiple Choice Questions Question 23

Answer:
13tan3x+C
Given:
sin2xcos4xdx
Hint:
You must know about the derivation of tan x and xndx
Explanation:
Let I=sin2xcos4xdx
=tan2xsec2xdx [sinxcosx=tanx;1cosx=secx]
Put tanx=t,sec2xdx=dt
=t2dt=(t)2+12+1+C=13tan3x+C
Hence,sin2xcos4x=13tan3x+C.

Indefinite Integrals Exercise Multiple Choice Questions Question 24

Answer:
ax+1xlogea
Given:
f(x)=(11x2)ax+1x,a>0
Hint:
You must know about the axdx
Explanation:
Let I=(11x2)ax+1xdx
=atdt  [Put x+1x=t(11x2)dx=dt]
=atlogea=ax+1xlogea [axdx=axloga]
Hence primitive of f(x) is ax+1xlogea

Indefinite Integrals Exercise Multiple Choice Questions Question 25

Answer:
log|1+logx|+C
Given:
1x+xlogxdx
Hint:
Using 1xdx
Explanation:
Let I=1x+xlogxdx
=1x(1+logx)dx=1x1+logxdx=dtt  [ Put 1+logx=t1xdx=dt]
=log|t|+C=log|1+logx|+C
Hence, 1x+xlogxdx=log|1+logx|+C

Indefinite Integrals Exercise Multiple Choice Questions Question 26

Answer:
sin1xx(1x)+C
Given:
x1xdx is equal to:
Hint:
You must know about the sin2θdθ . 
Explanation:
Let I =x1x
=sin2θ1sin2θ×2sinθcosθdθ  [ Put x=sin2θdx=2sinθcosθdθ]
=sin2θcos2θ2sinθcosθdθ [1sin2θ=cos2θ]
=tan2θ2sinθcosθdθ=sinθcosθ×2sinθcosθdθ=2sin2θdθ=(1cos2θ)dθ [2sin2θ=1cos2θ]
=1dθcos2θdθ=θ12sin2θ+C=θ12×2sinθcosθ+C=θsinθcosθ+C=θsinθ1sin2θ+C=sin1xx1x+C=sin1xx(1x)+C

Indefinite Integrals Exercise Multiple Choice Questions Question 27

Answer:
exf(x)+C
Given:
ex[f(x)+f(x)]dx
Hint:
You must know about the rule to integrate by parts
Explanation:
Let I=ex[f(x)+f(x)]dx
=exf(x)dx+exf(x)dx=f(x)exexf(x)dx+exf(x)dx{u.vdx=uvdxdudxvdxdx}=f(x)ex+C
Hence,ex[f(x)+f(x)]dx=exf(x)+C

Indefinite Integrals Exercise Multiple Choice Questions Question 28

Answer:
±log(sinxcosx)+C
Given:
sinx+cosx1sin2x
Hint:
You must know about the derivation of sin and cos function and 1xdx
Explanation:
Let I=sinx+cosx1sin2xdx
=sinx+cosxsin2x+cos2xsin2xdx [sin2x+cos2x=1]
=sinx+cosx(cosxsinx)2dx [(ab)2=a2+b22ab]
=sinx+cosx(cosxsinx)dx=dtt [ Put cosxsinx=t(cosx+sinx)dx=dt]
=log|t|+C=log(cosxsinx)+C
Hence,sinx+cosx1sin2x=log(cosxsinx)+C.

Indefinite Integrals Exercise Multiple Choice Questions Question 29

Answer:
sinx+C
Given:
xsinxdx=xcosx+α
Hint:
You must know the derivation of sin x and rule to integrate by parts
Explanation:
Let I=xsinxdx
=x(cosx)+(1)cosxdx  [ : :Integration by parts u.vdx=uvdxdudxvdxdx]
=xcosx+sinx+C
Acc to given
I=xcosx+αxcosx+sinx+C=xcosx+α
Comparing both sides
We get,α=sinx+C

Indefinite Integrals Exercise Multiple Choice Questions Question 30

Answer:
xtanx+C
Given:
cos2x1cos2x+1dx
Hint:
Using tan2xdx
Explanation:
Let I=cos2x1cos2x+1dx
=cos2xsin2x1cos2xsin2x+1dx [cos2x=cos2xsin2x]
=(cos2x1)sin2xcos2x+(1sin2x)dx=sin2xsin2xcos2x+cos2xdx [sin2x+cos2x=11sin2x=cos2x]
=2sin2x2cos2xdx=tan2xdx [tanx=sinxcosx]
=(sec2x1)dx=(1sec2x)dx=1dxsec2xdx=xtanx+C
Hence,cos2x1cos2x+1dx=xtanx+C

Indefinite Integrals Exercise Multiple Choice Questions Question 31

Answer:
2(sinx+xcosθ)+C
Given:
cos2xcos2θcosxcosθdx
Hint:
You must know about the integration of cos x
Explanation:
Let I=cos2xcos2θcosxcosθdx
=(2cos2x1)(2cos2θ1)cosxcosθdx [cos2θ=2cos2θ1]
=2cos2x2cos2θ+11cosxcosθdx=2cos2xcos2θcosxcosθdx=2(cosxcosθ)(cosx+cosθ)cosxcosθdx [a2b2=(a+b)(ab)]
=2cosxdx+2cosθdx=2cosxdx+2cosθ1dx=2sinx+2cosθx+C=2(sinx+xcosθ)+C
Hence,cos2xcos2θcosxcosθdx=2(sinx+xcosθ)+C.

Indefinite Integrals Exercise Multiple Choice Questions Question 32

Answer:
110[4+1x2]5+C
Given:
x9(4x2+1)6dx
Hint:
Using xndx
Explanation:
Let I=x9(4x2+1)6dx
=x9x12(4+1x2)6dx taking x2 common from denominator 
=1x3(4+1x2)6dx=1x3(4+1x2)6dx  [ Put 4+1x2=t2x3dx=dt]
=12dtt6=12[t6+16+1]+C=12[(t)55]+C=110×1t5+C=t510+C=110[4+1x2]5+C
Hence,x9(4x2+1)6dx=110[4+1x2]5+C

Indefinite Integrals Exercise Multiple Choice Questions Question 33

Answer:
a=13;b=1
Given:
x31+x2dx=a(1+x2)32+b1+x2+C
Hint:
Using xndx
Explanation:
Let I=x31+x2dx
=x2x1+x2dx=(t1)tdt2  [ Put 1+x2=tx2=t12xdx=dt]
=12t1tdt=12tdt121tdt=12(t)12dt12(t)12dt=12[t12+112+1]12[(t)12+112+1]+C=12×t323212×(t)1212+C=13t32t12+C
=13(1+x2)321+x2+C
Acc to given I=a(1+x2)82+b1+x2+C
13(1+x2)321+x2+C=a(1+x2)32+b1+x2+C
On comparing, we get
a=13,b=1

Indefinite Integrals Exercise Multiple Choice Questions Question 34

Answer:
xx22+x33log|1+x|+C
Given:
x3x+1dx
Hint:
You must know how to integrate xndxand 1xdx
Explanation:
Let I=x3x+1dx
=x3+11x+1dx=(x+1)(x2+1x)x+1dx1x+1dx [x3+1=(x+1)(x2+1x)]
=(x2+1x)dx11+xdx=x33+xx22log|1+x|+C=xx22+x33log|1+x|+C

Indefinite Integrals Exercise Multiple Choice Questions Question 35

Answer:
a=110,b=25
Given:
1(x+2)(x2+1)dx=alog|1+x2|+btan1x+15log|x+2|+C
Hint:
Using partial fraction and 1xdxand 11+x2dx
Explanation:
Let I=1(x+2)(x2+1)dx
Let 1(x+2)(x2+1)=Ax+2+Bx+Cx2+1 [Eq.(i))]
Multiplying by (x+2)(x2+1), we get
I=A(x2+1)+(Bx+C)(x+2)
Putting x = -2
1=A(5)+(2B+C)(2+2)1=5A+0A=15
Putting x = 01=A(1)+(B(0)+C)(0+2)1=15+2CC=25

Putting x = +1
1=A(1+1)+(B+C)(1+2)1=2(15)+(B+25)(3)1=25+3B+653B=1853B=35B=15
 From [Eq(i)]
1(x+2)(x2+1)=15x+2+(15)x+(25)x2+11(x+2)(x2+1)dx=151x+2dx+(15)122xx2+1dx+251x2+1dxI=15log|x+2|110log|1+x2|+25tan1x+C[Eq.(ii))]
Acc to given
I=alog|1+x2|+btan1x+15log|x+2| [Eq.(iii))]
 From Eq(ii) and Eq( iii ) we get a=110,b=25



Indefinite Integrals Exercise Multiple Choice Questions Question 36

Answer:
excosx+C
Given:
ex(cosxsinx)dx
Hint:
You must know the rule to integrate by parts
Explanation:
Let I=ex(cosxsinx)dx
=excosxdxexsinxdx=cosxex+exsinxdxexsinxdx..{uvdx=uvdxdudxvdxdx}=cosxex+C

Indefinite Integrals Exercise Multiple Choice Questions Question 37

Answer:
a=18,b=+78
Given:
3ex5ex4ex+5exdx=ax+bloge|4ex+5ex|+C
Hint:
Using exdx
Explanation:
We are given that
3ex5ex4ex+5exdx=ax+bloge|4ex+5ex|+C
Differentiate both sides
3ex5ex4ex+5ex=a+b1(4ex+5ex)(4ex5ex) [exdx=ex+C]
3ex5ex4ex+5ex=a(4ex+5ex)+b(4ex5ex)4ex+5ex3ex5ex=4aex+5aex+4bex5bex3ex5ex=(4a+4b)ex+(5a5b)ex
Comparing co eff of like terms, we get
4a+4b=3;5(ab)=5ab=1a=b14(b1)+4b=38b4=38b=7b=78a=781a=18

Indefinite Integrals Exercise Multiple Choice Questions Question 38

Answer:
ex(1+x2)+C
Given:
ex(1x1+x2)2dx
Hint:
Using integration by parts
Explanation:
Let I=ex(1x1+x2)2dx [(ab)2=a22ab+b2]
=ex(12x+x2)(1+x2)2dx
=ex(1+x2)(1+x2)2dxex2x(1+x2)2dx
=ex(1+x2)dxex2x(1+x2)2dx=ex11+x2dxex2x(1+x2)2dx
=11+x2exex((1+x2)(0)(1)(2x)(1+x2)2)dxex2x(1+x2)2dx{u.vdx=uvdxdudxvdxdx}
=ex1+x2+2xex(1+x2)2dx2xex(1+x2)2dx=ex1+x2+C

Indefinite Integrals Exercise Multiple Choice Questions Question 39

Answer:
xtanx2+C
Given:
x+sinx1+cosxdx
Hint:
Using integration by parts and sec2θdθ
Explanation:
Let I=x+sinx1+cosxdx
=x1+cosxdx+sinx1+cosxdx=x2cos2x2dx+2sinx2cosx2cos2x2dx [1+cos2θ=2cos2θ;sin2θ=2sinθcosθ]
=12xsec2x2dx+tanx2dx [ Put x2=tx=2tdx=2dt]
=122tsec2t.2dt+tant2dt=2tsec2tdt+2tantdt=2[t.tanttantdt]+2tantdt{u.vdx=uvdxdudxvdxdx}=2t.tant2tantdt+2tantdt=2ttant+C=2(x2)tanx2+C=xtanx2+C

Indefinite Integrals Exercise Multiple Choice Questions Question 40

Answer:
(x+1)tan1xx+C
Given:
tan1xdx
Hint:
Using integration by parts and 11+x2dx
Explanation:
Let I=tan1xdx
I=tan1t.2tdt [ Put x=t12xdx=dtdx=2xdtdx=2tdt]
=2tan1ttdt
=2[tan1tt2211+t2t22dt]{u.vdx=uvdxdudxvdxdx}
=2tan1tt22(t2+11t2+1)dt=tan1tt21dt+11+t2dt=t2tan1tt+tan1t+C=xtan1xx+tan1x+C=(x+1)tan1xx+C

Indefinite Integrals Exercise Multiple Choice Questions Question 41

Answer:
cosec(ba)log|sin(xb)sin(xa)|
Given:
1sin(xa)sin(xb)dx
Hint:
Using sin(xy)=sinxcosycosxsiny and 1xdx
Explanation:
Let I=1sin(xa)sin(xb)dx
=1sin(ba)sin(ba)sin(xa)sin(xb)dx multiplying by sin(ba) in num.\& den. 
=1sin(ba)sin((xa)(xb))sin(xa)sin(xb)dx [sin(xy)=sinxcosycosxsiny]
=1sin(ba)sin(xa)cos(xb)cos(xa)sin(xb)sin(xa)sin(xb)dx=1sin(ba)[cos(xb)sin(xb)dxcos(xa)sin(xa)]dx=1sin(ba)[log|sin(xb)|log|sin(xa)|]+C
=cosec(ba)log|sin(xb)sin(xa)| [logalogb=logab]

Indefinite Integrals Exercise Multiple Choice Questions Question 42

Answer:
13ex3+C
Given:
x2ex3dx
Hint:
Using exdx
Explanation:
Let I=x2ex3dx
=etdt3  [Put x3=t3x2dx=dtx2dx=dt3]
=13et+C=13ex3+C [exdx=ex+C]

Indefinite Integrals Exercise Multiple Choice Questions Question 43

Answer:
exlog(x+1)+C
Given:
exx+1[1+(x+1)log(x+1)]dx
Hint:
Using integration by parts and exdx
Explanation:
Let I=exx+1[1+(x+1)log(x+1)]dx
=ex[1x+1+log(x+1)]dx=exlog(x+1)dx+ex1x+1dx
=log(x+1)ex1x+1exdx+1x+1exdx{uvdx=uvdxdudxvdxdx}
=log(x+1)ex+C
Hence,exx+1[1+(x+1)log(x+1)]dx=log(x+1)ex+C

In this class 12 RD Sharma chapter 18 exercise MCQ. At Career360 Class 12 Maths, RD Sharma Class 12 Solutions Indefinite Integrals Ex MCQ helps students get a fair academic score in the exam.

This chapter of RD Sharma class 12th exercise MCQ Indefinite Integrals primary based on the possibility of soundness. Students can download the RD Sharma class 12 solutions MCQ Chapter 18 Indefinite Integrals to look into this theme.

Some important concepts like:-

  • Definition of primitive or antiderivative

  • Questions related to Fundamental integration formulae

  • Questions related to Integration of trigonometric functions

  • Geometrical interpretation of indefinite integral

  • Questions related to Comparison between differentiation and integration

  • Methods of integration

  • Questions related to Integration by substitution

  • Questions related to Integration by parts

  • Questions related to some important integrals along with theorems

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