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RD Sharma Class 12 Exercise 18.16 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.16 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:16 PM IST

Countless students in India have chosen the RD Sharma class 12th exercise 18.16 solution to be their holy grail when it comes to exam preparations. It cannot be argued that RD Sharma Solutions have made a name for themselves among other NCERT solutions. The answers provided in the RD Sharma class 12 chapter 18 exercise 18.16 deserves special mention because the chapter is extremely important for students in class 12.

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
  2. Indefinite Integrals Excercise:18.16
  3. RD Sharma Chapter wise Solutions

The RD Sharma class 12 solutions Indefinite Integrals ex 18.16 is arguably the best NCERT solution in the RD Sharma solutions series which deals with chapter 18 of the NCERT textbook. Students will find in this chapter concepts of Integrals, Graphs of indefinite integrals, Indefinite integrals of common functions,etc. Exercise 18.16 has 16 questions which tests students on basic concepts of the chapter. The RD Sharma class 12th exercise 18.16 solutions will help them solve these problems and get accurate answers.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.16

Indefinite Integrals exercise 18.16 question 1

Answer:
12log|1+tanx1tanx|+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
sec2x1tan2xdx
Solution:
Letsec2x1tan2xdx
Puttanx=tsec2xdx=dt
ThenI=11t2dt=112t2dt
=12×1log|1+t1t|+C[1a2x2dx=12alog|a+xax|+C]=12log|1+tanx1tanx|+C[t=tanx]

Indefinite Integrals exercise 18.16 question 2

Answer:
tan1(ex)+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
ex1+e2xdx
Solution:
LetI=ex1+e2xdx
Putex=texdx=dt
ThenI=11+t2dx
=tan1(t1)+C[1a2+x2dx=1atan1(xa)+C]=tan1(ex)+C[t=ex]

Indefinite Integrals exercise 18.16 question 3

Answer:
tan1(sinx+2)+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
cosxsin2x+4sinx+5dx
Solution:
LetI=cosxsin2x+4sinx+5dx
Putsinx=tcosxdx=dt
ThenI=1t2+4t+5dx
=1t2+22t+2222+5dt=1(t+2)24+5dt=1(t+2)2+1dt
Putt+2=udt=du
ThenI=1u2+1du
=tan1(u1)+C[1a2+x2dx=1atan1(xa)+C]=tan1(t+21)+C[u=t+2]=tan1(sinx+2)+C[t=sinx]

Indefinite Integrals exercise 18.16 question 4

Answer:
log|ex+2ex+3|+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
exe2x+5ex+6dx
Solution:
LetI=exe2x+5ex+6dx
Putex=texdx=dt
ThenI=1t2+5t+6dt
=1t2+2t52+(52)2(52)2+6dt=1(t+52)2254+6dt=1(t+52)225+244dt
=1(t+52)214dt=1(t+52)2(12)2dt
Putt+52=udt=du
=1u2(12)2du
=12×12log|u12u+12|+C[1x2a2dx=12alog|xax+a|+C]=logt+5212t+5212+C
=log|2t+5122t+5+12|+C[u=t+52]
. I=log|2t+42t+6|+C=log|t+2t+3|+CI=log|ex+2ex+3|+C[t=ex]

Indefinite Integrals exercise 18.16 question 5

Answer:
136log|2e3x32e3x+3|+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
e3x4e6x9dx
Solution:
Lete3x4e6x9dx
Pute3x=t3e3xdx=dte3xdx=dt3
ThenI=14t29dt3=14(t294)dt3
=14×31t294dt=14×31t2(32)2dt
=14×3[12×32log|t32t+32|]+C[1x2a2dx=12alog|xax+a|+C]
=14×3×3log|2t322t+32|+C=136log|2t32t+3|+C=136log|2e3x32e3x+3|+C[t=e3x]

Indefinite Integrals exercise 18.16 question 6

Answer:
tan1(ex)+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
1ex+exdx
Solution:
LetI=1ex+exdx=1ex+1exdx
=1e2x+1exdx=exe2x+1dx
Putex=texdx=dt
ThenI=1t2+1dt
=tan1(t)+C[1x2+a2dx=1atan1(xa)+C]=tan1(ex)+C[ex=t]

Indefinite Integrals exercise 18.16 question 7

Answer:
122tan1(x2+12)+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
xx4+2x2+3dx
Solution:
LetI=xx4+2x2+3dx
Putx2=t2xdx=dtxdx=dt2
ThenI=1t2+2t+3dt2
=121t2+2t+3dt=121t2+2t1+1212+3dt=121(t+1)21+3dt=121(t+1)2+2dt=121(t+1)2+(2)2dt
Putu=t+1du=dt
ThenI=121u2+(2)2du
=12×12tan1(u2)+C[1x2+a2dx=1atan1(xa)+C]=122tan1(t+12)+C[u=t+1]=122tan1(x2+12)+C[t=x2]




Indefinite Integrals exercise 18.16 question 8

Answer:
12tan1(x6)+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
3x51+x12dx
Solution:
LetI=3x51+x12dx
=3x51+(x6)2dx
Putx6=t6x5dx=dt
ThenI=3x51+t2dt6x5
=1211+t2dt=12112+t2dt=1211tan1(t1)+C[1a2+x2dx=1atan1(xa)+C]=12tan1(t)+C=12tan1(x6)+C[t=x6]

Indefinite Integrals exercise 18.16 question 9

Answer:
16a3log|x3a3x3+a3|+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
x2x6a6dx
Solution:
LetI=x2x6a6dx
=x2(x3)2(a3)2dx
Putx3=t3x3dx=dtx2dx=dt3
ThenI=1t2(a3)2dt2
=131t2(a3)2dt=13×12(a3)log|ta3t+a3|+C[1x2a2dx=12alog|xax+a|+C]=16a3log|x3a3x3+a3|+C[t=x3]

Indefinite Integrals exercise 18.16 question 10

Answer:
13a3tan1(x3a3)+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
x2x6+a6dx
Solution:
LetI=x2x6+a6dx
=x2(x3)2+(a3)2dx
Putx3=t3x2dx=dtx2dx=dt3
ThenI=1t2+(a3)2dt
=131t2+(a3)2dt=13×1a3tan1(ta3)+C[1a2+x2dx=1atan1(xa)+C]=13a3tan1(x3a3)+C[t=x3]

Indefinite Integrals exercise 18.16 question 11

Answer:
16log|x6x6+1|+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
1x(x6+1)dx
Solution:
LetI=1x(x6+1)dx
=x5x5{x(x6+1)}dx=x5x6(x6+1)dx
Putx6=t6x5dx=dtx5dx=dt6
ThenI=1t(t+1)dt6
=161t2+tdt=161t2+2t12+(12)2(12)2dt=161(t+12)214dt
Putt+12=udt=du
=161u2(12)2dt=161u2(12)2du
=1612×(12)log|u12u+12|+C[1x2a2dx=12alog|xax+a|+C]=16log|2u122u+12|+C=16log|2u12u+1|+C
=16log|2(t+12)12(t+12)+1|+C[t+12=u]=16log|2t+112t+1+1|+C=16log|2t2t+2|+C=16log|tt+1|+C=16log|x6x6+1|+C[t=x6]

Indefinite Integrals exercise 18.16 question 12

Answer:
13tan1(2x213)+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
xx4x2+1dx
Solution:
LetI=xx4x2+1dx
=x(x2)2x2+1dx
Putx2=t2xdx=dtxdx=dt2
=1t2t+1dt2
=121t22t12+(12)2(12)2+1dt=121(t12)214+1dt=121(t12)2(144)dt=121(t12)2(34)dt=121(t12)2+34dt
Lett12=udt=du
ThenI=121u2+(32)2du
=12×1(32)tan1(u32)+C[1a2+x2dx=1atan1(xa)+C]=13tan1(t1232)+C[u=t12]=13tan1(2t1232)+C=13tan1(2t13)+C=13tan1(2x213)+C[t=x2]

Indefinite Integrals exercise 18.16 question 13

Answer:
348log|x2343x23+43|+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
x3x418x2+11dx
Solution:
LetI=x3x418x2+11dx
=x3(x2)218x2+11dx
Putx2=t2xdx=dtxdx=dt2
ThenI=13t218t+11dt2
=1213(t26t+113)dt=161t22.3.t+(3)2(3)2+113dt=161(t3)29+113dt=161(t3)2(27113)dt=161(t3)2163dt=161(t3)2(43)2dt
Putt3=udt=du
ThenI=161u2(43)2du
=161243log|u43u+43|+C[1x2a2dx=12alog|xax+a|+C]=36×8log|u43u+43|+C=348log|t343t3+43|+C[u=t3]
=348log|x2343x23+43|+C[t=x2]

Indefinite Integrals exercise 18.16 question 14

Answer:
log|ex+1ex+2|+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
GIven:
ex(1+ex)(2+ex)dx
Solution:
LetI=ex(1+ex)(2+ex)dx
Letex=texdx=dt
ThenI=1(1+t)(2+t)dt
=1t2+2t+t+2dt=1t2+3t+2dt=1t2+2t32+(32)2(32)2+2dt=1(t+32)294+2dt=1(t+32)2(984)dt=1(t+32)214dt=1(t+32)2(12)2dt
Putt+32=udt=du
ThenI=1u2(12)2du
=12×12log|u12u+12|+C[1x2a2dx=12alog|xax+a|+C]=log|t+3212t+32+12|+C[t+32=u]=log|2t+22t+4|+C=log|t+1t+2|+C
=log|ex+1ex+2|+C[t=ex]

Indefinite Integrals exercise 18.16 question 15

Answer:
16log|1+tan3x1tan3x|+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
tan2xsec2x1tan6xdx
Solution:
LetI=tan2xsec2x1tan6xdx
=tan2xsec2x1(tan3x)2dx
 Put tan3x=t3tan2xsec2xdx=dt Then I=11t2dt3=13112t2dt=1312×1log|1+t1t|+C[1a2x2dx=12alog|a+xax|+C]=16log|1+tan3x1tan3x|+C[t=tan3x]

Indefinite Integrals exercise 18.16 question 16

Answer:
123log|3+sinxcosx3sinx+cosx|tan1(sinx+cosx)+C
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
1cosx+cosecxdx
Solution:
LetI=1cosx+cosecxdx
=1cosx+1sinxdx[cosecx=1sinx]=1cosxsinx+1sinxdx=sinxcosxsinx+1dx=2sinx2(cosxsinx+1)dx=sinx+sinx2+2sinxcosxdx=sinx+cosxcosx+sinx2+2sinxcosxdx={sinx+cosx2+2sinxcosx+sinxcosx2+2sinxcosx}dx
=sinx+cosx2+2sinxcosxdx+sinxcosx2+2sinxcosxdx=sinx+cosx31+2sinxcosxdx+sinxcosx1+1+2sinxcosxdx=sinx+cosx3(sin2x+cos2x)+2sinxcosxdx+sinxcosx1+(sinx+cos2x)+2sinxcosxdx[1=sin2x+cos2x]=sinx+cosx3sin2xcos2x+2sinxcosxdx+sinxcosx1+(sin2x+cos2x+2sinxcosx)dx
=sinx+cosx3(sin2x+cos2x2sinxcosx)dx+sinxcosx1+(sinx+cosx)2dx=sinx+cosx3(sinxcosx)2dx+sinxcosx1+(sinx+cosx)2dx Let I=I1+I2...(i)1cosx+cosecxdx=sinx+cosx3(sinxcosx)2dx+sinxcosx1+(sinx+cosx)2dx Where I1=sinx+cosx3(sinxcosx)2dx
 And I2=sinxcosx1+(sinx+cosx)2dx
Here first we find the integral of I1 and I2 then we put the value of I1 and I2 in I and then we get the required solution
 So I1=sinx+cosx3(sinxcosx)2dx Put t=sinxcosxdt=[cosx(sinx)]dxdt=[cosx+sinx]dx Then I1=13t2dt=1(3)2t2dt=123log|3+t3t|+C1[1a2x2dx=12alog|a+xax|+C]=123log|3+sinxcosx3(sinxcosx)|+C1[t=sinxcosx]=123log|3+sinxcosx3sinx+cosx|+C1......(ii) And I2=sinxcosx1+(sinx+cosx)2dx Put u=sinx+cosxdu=(cosxsinx)dxdu=(sinxcosx)dx(sinxcosx)dx=duI2=11+u2(du)=112+u2du=tan1(u1)+C2[1a2+x2dx=1atan1(xa)+C]=tan1(sinx+cosx)+C2..... (iii) [u=sinx+cosx] Putting the value of (ii) and (iii) in (i) we get I=I1+I2=123log|3+sinxcosx3sinx+cosx|+C1+{tan1(sinx+cosx)+C2}=123log|3+sinxcosx3sinx+cosx|tan1(sinx+cosx)+C[C=C1+C2]

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