#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 61

$\frac{1}{6} \log (\cos x-1)+\frac{1}{2} \log (\cos x+1)-\frac{2}{3} \log \left(\cos x+\frac{1}{2}\right)+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{\sin x+\sin 2 x} d x \\$

Explanation:

Let

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

Put

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

$I=\int \frac{1}{2\left(t^{2}-1\right)\left(\frac{1}{2}+t\right)} d t$                            (1)

$\frac{1}{(t-1)(t+1)\left(\frac{1}{2}+t\right)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{\frac{1}{2}+t}$                          (2)

$1=A(t+1)\left(\frac{1}{2}+t\right)+B(t-1)\left(\frac{1}{2}+t\right)+C\left(t^{2}-1\right)$

At $t=- 1$

\begin{aligned} &1=-2 B\left(\frac{-1}{2}\right) \\ &B=1 \end{aligned}

At $t= 1$

\begin{aligned} &1=2 A\left(\frac{3}{2}\right) \\ &A=\frac{1}{3} \end{aligned}

\begin{aligned} &\text { At } t=-\frac{1}{2} \\ &1=C\left(\frac{1}{4}-1\right) \\ &1=C\left(\frac{-3}{4}\right) \\ &C=\frac{-4}{3} \end{aligned}

Put in (1) using (2)
\begin{aligned} &I=\frac{1}{2} \cdot \frac{1}{3} \int \frac{1}{t-1} d t+\frac{1}{2} \int \frac{1}{t+1} d t-\frac{1}{2} \cdot \frac{4}{3} \int \frac{1}{t+\frac{1}{2}} d t \\ &I=\frac{1}{6} \log (t-1)+\frac{1}{2} \log (t+1)-\frac{2}{3} \log \left(t+\frac{1}{2}\right)+C \\ &I=\frac{1}{6} \log (\cos x-1)+\frac{1}{2} \log (\cos x+1)-\frac{2}{3} \log \left(\cos x+\frac{1}{2}\right)+C \end{aligned}