Question

Evaluate $\int \frac{1}{\sin x+\sin 2x}dx$

Answer 1

$\int \frac{1}{\sin x+\sin 2x}dx$

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

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

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

put $1+2\cos x=t,\cos x=\frac{t-1}{2}-2\sin xdx=dt$

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

$=-\frac{1}{2}\int \frac{1}{(1+t)(3-t)t}dt=\frac{A}{1+t}+\frac{B}{3-t}+\frac{C}{t}$

$=A=\frac{-1}{48},B=\frac{1}{6},C=\frac{-1}{12}$

$\Rightarrow -\int \frac{1}{2(1+t)(3-t)t}dt=\int \frac{-1}{48(3-t)}dt+\int \frac{1}{6(1+t)}dt-\int \frac{1}{12t}dt$

$=\frac{-1}{48}ln|3-t|+\frac{1}{6}ln|1+t|\frac{-1}{12}ln|t|+C$

$=-\frac{-1}{48}ln|2-2\cos x|+\frac{1}{6}ln|2+2\cos x|\frac{-1}{12}ln|1+2\cos x|+C$