Question

Evaluate : $\int \frac{dx}{\sin x+\sin 2x}$.

Answer 1

Let $I=\int \frac{dx}{\sin x+\sin 2x}=\int \frac{dx}{2\sin x\cos x+\sin x}$

$=\int \frac{dx}{\sin x(1+2\cos x)}$

Take $\cos x=u$

$\Rightarrow \cos x=u\Rightarrow -\sin xdx=du$

$\frac{dx}{\sin x}=\frac{-du}{{\sin }^{2}x}$ (divide by both ${\sin }^{2}x$)

$\Rightarrow \frac{dx}{\sin x}=-\frac{du}{(1-{\cos }^{2}x)}=\frac{du}{{\cos }^{2}x-1}$

$\therefore I=\int \frac{du}{(u^2-1)(2u+1)}=\int \frac{1}{(u-1)(u+1)(2u+1)}$

$\frac{1}{(u-1)(u+1)(2u+1)}=\frac{A}{u-1}+\frac{B}{u+1}+\frac{C}{2u+1}$

$1=A(u+1)(2u+1)+B(u-1)(2u+1)+C(u-1)(u+1)$ ...(i)

Solving $A,B$ and $C$

we get $A=\frac{1}{6},B=\frac{1}{2},C=\frac{-4}{3}$

$\therefore I=\frac{1}{6}\int \frac{du}{u-1}+\frac{1}{2}\int \frac{du}{u+1}+\frac{-4}{3}\int \frac{1}{2u+1}$

$=\frac{1}{6}\ln (u-1)+\frac{1}{2}\ln (u+1)-\frac{4}{3}\frac{\ln (2u+1)}{2}$

$=\frac{1}{6}\ln (\cos x-1)\frac{1}{2}\ln (\cos x+1)-\frac{2}{3}\ln (2\cos x+1)+C$