Question

Solve $\int \frac{1}{\sin \theta +\sin 2\theta }d\theta$

Answer 1

Let $I=\int \frac{1}{\sin \theta +\sin 2\theta }d\theta$

$\Rightarrow I=\int \frac{1}{\sin 2\theta +\sin 2\theta }d\theta$

$\Rightarrow I=\int \frac{1}{\sin \theta +2\sin \theta \cos \theta }d\theta$

$\Rightarrow I=\int \frac{1}{\sin \theta \left(1+2\cos \theta \right)}d\theta$

$\Rightarrow I=\int \frac{1\times \sin \theta d\theta }{\sin \theta \times \sin \theta \left(1+2\cos \theta \right)}$

$\Rightarrow I=\int \frac{\sin \theta d\theta }{{\sin }^2\theta \left(1+2\cos \theta \right)}$

$\Rightarrow I=\int \frac{\sin \theta d\theta }{\left(1-{\cos }^2\theta \right)\left(1+2\cos \theta \right)}$

$\Rightarrow I=\int \frac{\sin \theta d\theta }{\left(1+\cos \theta \right)\left(1-\cos \theta \right)\left(1+2\cos \theta \right)}$

Put $t=\cos \theta$

$\frac{dt}{d\theta }=-\sin \theta$

$\Rightarrow -dt=\sin \theta d\theta$

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

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

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

On solving

$A=\frac{1}{6}$

$B=-\frac{1}{2}$

$C=\frac{4}{5}$

$I=\int \frac{A}{1-t}dt+\int \frac{B}{1+t}dt+\int \frac{C}{1+2t}dt$

$=\int \frac{\frac{1}{6}}{1-t}dt+\int \frac{\frac{-1}{2}}{1+t}dt+\int \frac{\frac{4}{5}}{1+2t}dt$

$=\frac{1}{6}\log \left(1-t\right)-\frac{1}{2}\log \left(1+t\right)+\frac{4}{5}\frac{\log \left(1+2t\right)}{2}+C$

$\frac{-1}{6}\log \left|\cos \theta \right|-\frac{1}{2}\log \left|1+\cos \theta \right|+\frac{2}{3}\log \left|1+2\cos \theta \right|+C$

Hence,

$I=\int \frac{1}{\sin \theta +\sin 2\theta }d\theta$

$=-\left[\frac{-1}{6}\log \left|1-cos\theta \right|-\frac{1}{2}\log \left|1+\cos \theta \right|+\frac{2}{3}\log \left|1+2\cos \theta \right|+C\right]$

$=\frac{1}{6}\log \left|1-\cos \theta \right|+\frac{1}{2}\log \left|1+\cos \theta \right|-\frac{2}{3}\log \left|1+2\cos \theta \right|+C$