$\int \frac{d x}{\sin x+\sin 2 x}=$
$\int \frac{d x}{\sin x+\sin 2 x}=$
- $\begin{array}{r}\frac{1}{2} \log _e|1+\cos x|+\frac{1}{6} \log _e|1-\cos x|-\frac{2}{3} \log _e \mid \\ 1+2 \cos x \mid+c\end{array}$
- $\begin{array}{r}\frac{1}{2} \log _e|1+\cos x|-\frac{2}{3} \log _e|1-\cos x|+\frac{1}{2} \log _e \mid \\ 1+2 \cos x \mid+c\end{array}$
- $\begin{array}{r}\frac{1}{2} \log _e|1+\sin x|-\frac{1}{3} \log _e|1-\sin x|-\frac{1}{3} \log _e \mid \\ 1+\cos x \mid+c\end{array}$
- $\begin{array}{r}\frac{1}{3} \log _e|1-\sin x|+\frac{1}{2} \log _e|1+\cos x|-\frac{2}{3} \log _e \mid \\ 1-2 \cos x \mid+c\end{array}$
Solution
Given,
$
\begin{aligned}
& \int \frac{d x}{\sin x+\sin 2 x}=\int \frac{d x}{\sin x+2 \sin x \cos 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}{\left(\cos ^2 x-1\right)(1+2 \cos x)}
\end{aligned}
$
Let $\cos x=t$
$
\Rightarrow-\sin x d x=d t=\int \frac{d t}{\left(t^2-1\right)(1+2 t)}
$
By partial fraction,
$
\begin{aligned}
& \frac{1}{(t-1)(t+1)(2 t+1)}=\frac{A}{(t-1)}+\frac{B}{(t+1)}+\frac{C}{(2 t+1)} \\
& \Rightarrow \quad 1=A(t+1)(2 t+1)+B(t-1)(2 t+1)+C\left(t^2-1\right) \\
& \text { At } \quad t=-1 \\
& \Rightarrow \quad 1=A(0)+B(-2)(-1)+C(0) \\
& \Rightarrow \quad 1=2 B \Rightarrow B=\frac{1}{2}
\end{aligned}
$
At $t=-\frac{1}{2} \Rightarrow 1=A(0)+B(0)+C\left(-\frac{3}{4}\right)$
$
\begin{aligned}
& \Rightarrow \quad C=-\frac{4}{3} \\
& \text { At } t=0, \\
& \Rightarrow 1=A-B-C \Rightarrow 1=A-\frac{1}{2}+\frac{4}{3}
\end{aligned}
$
$\begin{aligned} & \Rightarrow A=1+\frac{1}{2}-\frac{4}{3}=\frac{1}{6} \\ & \therefore A=\frac{1}{6}, B=\frac{1}{2} \text { and } C=-\frac{4}{3} \\ & \text { Now, } \int \frac{d t}{\left(t^2-1\right)(1+2 t)}=\int \frac{1 / 6}{(t-1)} d t+\frac{1}{2} \int \frac{d t}{t+1} \\ & -\frac{4}{6} \int \frac{d t}{\left(t+\frac{1}{2}\right)} \\ & =\frac{\ln (t-1)}{6}+\frac{1}{2} \ln (t+1)-\frac{4}{6} \ln \left(t+\frac{1}{2}\right)+C \\ & =\frac{\ln (\cos x-1)}{6}+\frac{1}{2} \ln (\cos x+1)-\frac{2}{3} \ln \left(\cos x+\frac{1}{2}\right)+C \\ & =\frac{1}{2} \ln |(1+\cos x)|+\left|\frac{1}{6} \ln \right|(1-\cos x) \mid \\ & -\frac{2}{3} \ln \left|\cos x+\frac{1}{2}\right|+C \\ & =\frac{1}{2} \ln |(1+\cos x)|+\frac{1}{6} \ln \mid(1-\cos x \mid \\ & \frac{-2}{3} \ln |(1+2 \cos x)|+C \\ & \end{aligned}$
Asked in: AP EAMCET 2019 (21 Apr Shift 1)
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