Question

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

Answer 1

$I=\int \frac{1dx}{sinx+sin2x}$

$\therefore I=\int \frac{1dx}{sinx+2sinxcosx}$

$\therefore I=\int \frac{1dx}{sinx(1+2cosx)}$

$\therefore I=\int \frac{sinxdx}{si{n}^{2}x(1+2cosx)}$

$\therefore I=\int \frac{-sindx}{(cosx+1)(cosx-1)(2cosx+1)}$

Let u= cos x du=-sin x dx

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

partial fraction problem

A(u-1)(2u+1)+B(u+1)(2u+1)+C(u-1)(u+1)=a

$\therefore A=\frac{1}{2}B=\frac{1}{6}C=-\frac{4}{3}$

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

$-\frac{2}{3}\int \frac{2}{2u+1}du$

$\therefore I=\frac{1}{2}ln|u+1)+\frac{1}{6}ln|4-1)-\frac{2}{3}ln|2u+1)+c$