# Get Answers to all your Questions

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

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

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}