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 67 maths textbook solution.

Please solve RD Sharma Class 12 Chapter Indefinite Integral Exercise Rivision Exercise Question 67 maths textbook solution.

Answers (1)

Answer :

=\frac{1}{6} \ln (1-\cos x)+\frac{1}{2} \ln (1+\cos x)-\frac{2}{3} \ln (1+2 \cos x)+c

Hint  : To solve the given statement we will use partial fraction rule.

Given :  \int \frac{1}{\sin x+\sin 2 x} d x

Solution :

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

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

\begin{aligned} &1=A+2 A t+A t+2 A t^{2}+B+2 B t-B t-2 B t^{2}+C-C t^{2} \\ &0=2 A-2 B-C \ldots \ldots \ldots . . (1)\end{aligned}

\begin{aligned} &0=3 A+B \ldots \ldots \ldots \ldots \text { (2) } \quad B=-3 A \\ &1=A+B+C \ldots \ldots \ldots \text { (3) } \end{aligned}

On solving, A=\frac{1}{6}, B=-\frac{1}{2}, C=\frac{4}{3}

\int \frac{1}{(1-t)(1+t)(1+2 t)} d t=-\int \frac{1}{6(1-t)}+\int \frac{1}{2(1+t)}-\int \frac{4}{3} \frac{1}{(1+2 t)} d t

\begin{aligned} &=-\frac{1}{6}-\ln (1-t)+\frac{1}{2} \ln (1+t)-\frac{4}{3} \times \frac{1}{2} \ln (1+2 t) \\ &=\frac{1}{6} \ln (1-\cos x)+\frac{1}{2} \ln (1+\cos x)-\frac{2}{3} \ln (1+2 \cos x)+c \end{aligned}

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 practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question