$$ \int \frac{d x}{\sin x+\sin 2 x}= $$
$$
\int \frac{d x}{\sin x+\sin 2 x}=
$$
- $\frac{1}{6} \log (1-\cos x)+\frac{1}{2} \log (1+\cos x)+$
$$
\frac{2}{3} \log |1+2 \cos x|+c
$$
- $$
\begin{aligned}
& \frac{1}{6} \log (1-\cos x)-\frac{1}{2} \log (1+\cos x)- \\
& \frac{2}{3} \log |1+2 \cos x|+c
\end{aligned}
$$
- $$
\begin{aligned}
& \frac{1}{6} \log (1-\cos x)+\frac{1}{2} \log (1+\cos x)- \\
& \frac{2}{3} \log |1+2 \cos x|+c
\end{aligned}
$$
- $\frac{1}{6} \log [(1-\cos x)(1+\cos x)|1+2 \cos x|]+c$
Solution
$
\begin{aligned}
& \text { } I=\int \frac{d x}{\sin x+\sin 2 x} \\
& =\int \frac{d x}{\sin x(1+2 \cos x)}=\int \frac{\sin x d x}{\sin ^2 x(1+2 \cos x)} \\
& =\int \frac{\sin x d x}{(1-\cos x)(1+\cos x)(1+2 \cos x)}
\end{aligned}
$
Let $\cos x=t \Rightarrow-\sin x d x=d t$
So,
$
I=-\int \frac{d t}{(1-t)(1+t)(1+2 t)}
$
By partial fraction method
$
\begin{aligned}
& \frac{1}{(1-t)(1+t)(1+2 t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1+2 t} \\
& \Rightarrow 1=A(1+t)(1+2 t)+B(1-t) \\
& A=\frac{1}{6}, B=\frac{-1}{2} \text { and } C=\frac{4}{3}
\end{aligned}
$
$A=\frac{1}{6}, B=\frac{-1}{2} \quad$ and $C=\frac{4}{3}$
So,
$
\begin{aligned}
I=-\frac{1}{6} \int \frac{d t}{1-t} & +\frac{1}{2} \int \frac{d t}{1+t}-\frac{4}{3} \int \frac{d t}{1+2 t} \\
=\frac{1}{6} \log (1-\cos x) & +\frac{1}{2} \log (1+\cos x) \\
& -\frac{2}{3} \log (1+2 \cos x)+c .
\end{aligned}
$
Asked in: AP EAMCET 2018 (23 Apr Shift 2)
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