Question

Evaluate: $\int \frac{d\theta }{\sin \theta +\sin 2\theta }$

Answer 1

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

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

Multiplying Numerator and denominator by $\tan \theta {\sec }^2\theta$, we get

$I=\int \frac{\tan \theta {\sec }^2\theta }{2{\tan }^2\theta +\sec \theta {\tan }^2\theta }d\theta$

Assuming $\sec \theta =x\Rightarrow \sec \theta \tan \theta =dx$

$I=\int \frac{xdx}{2(x^2-1)+x(x^2-1)}$

$=\int \frac{xdx}{(2+x)(x^2-1)}$

Using Partial fractions, we get

$I=\int (\frac{1}{2(x+1)}-\frac{2}{3(x+2)}+\frac{1}{6(x-1)})dx$

$\Rightarrow I=\frac{1}{2}\log (x+1)-\frac{2}{3}\log (x+2)+\frac{1}{6}\log (x-1)+C$