Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma Class 12 Chapter Indefinite Integral Exercise Rivision Exercise Question 66 maths textbook solution.

Please solve RD Sharma Class 12 Chapter Indefinite Integral Exercise Rivision Exercise Question 66 maths textbook solution.

Answers (1)

Answer :

I=+\frac{1}{2} \ln |1+\cos x|-\frac{1}{10} \ln |1-\cos x|-\frac{3}{5} \ln |3 \cos x+2|+c

Hint: To solve the given solution we multiply and divide the given statement with sin x.

Given :  \int \frac{1}{\sin x(2+3 \cos x)} d x

Solution : 

\begin{aligned} &I=\int \frac{1}{\sin x(2+3 \cos x)} d x \\ &I=\int \frac{\sin x}{\sin ^{2} x(2+3 \cos x)} d x \\ &I=\int \frac{\sin x d x}{\left(1-\cos ^{2} x\right)(2+3 \cos x)} \end{aligned}

\begin{aligned} &t=\cos x, d t=-\sin x d x \\ &I=\int \frac{-d t}{\left(1-t^{2}\right)(2+3 t)} \\ &I=\int \frac{-d t}{(1+t)(1-t)(2+3 t)} \end{aligned}

\\\frac{1}{(1+t)(1-t)(2-3 t)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{c}{2+3 t}=A(1-t)(2+3 t)+B(1+t)(2+3 t)+C(1+t)(1-t)

\begin{array}{ll} A=\frac{1}{(1-(-1))(2-3)}=-\frac{1}{2} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because 1+t=0, t=-1]} \\ B=\frac{1}{(1+1)(2+3)}=\frac{1}{10} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because 1-t=0, t=1]} \\ C=\frac{1}{1-\left(\frac{4}{9}\right)}=\frac{9}{5} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\because 2+3 t=0, t=-\frac{2}{3}\right]} \end{array}

\begin{aligned} &I=-\int d t\left(\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{2+3 t}\right) \\ &I=-\left[A \int \frac{1}{1+t} d t+B \int \frac{1}{1-t} d t+C \int \frac{1}{2+3 t} d t\right. \end{aligned}

\begin{aligned} &I=-\left[-\frac{1}{2} \ln |1+t|+\frac{1}{10} \ln |1-t|+\frac{9}{5} \times \frac{1}{3} \ln |2+3 t|+c\right. \\ &I=+\frac{1}{2} \ln |1+t|-\frac{1}{10} \ln |1-t|-\frac{3}{5} \ln |3 t+2|+c \end{aligned}

I=+\frac{1}{2} \ln |1+\cos x|-\frac{1}{10} \ln |1-\cos x|-\frac{3}{5} \ln |3 \cos x+2|+c

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Date Sheet 2027 Live: Class 10, 12 exam dates soon; LOC submission process, paper pattern
anam-khan

CBSE Exam Date 2027 Highlights: Class 10, 12 exam dates soon on cbse.gov.in; private candidate form
anam-khan

CBSE 10th, 12th board exam schedule 2027: Sample papers, passing marks

Latest Question