Question

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

Answer 1

We have,
$$I=\int \frac{dx}{\sin x+\sin 2x}$$
$$=\int \frac{dx}{\sin x+2\sin x\cos x}$$
$$=\int \frac{dx}{\sin x(1+2\cos x)}$$
$$=\int \frac{\sin xdx}{{\sin }^{2}x(1+2\cos x)}$$
$$=\int \frac{\sin xdx}{(1-{\cos }^{2}x)(1+2\cos x)}$$
$$=\int \frac{\sin xdx}{(1-\cos x)(1+\cos x)(1+2\cos x)}$$
$$\text{Putting }\cos x=t$$
$$\Rightarrow -\sin xdx=dt$$
$$\Rightarrow \sin xdx=-dt$$
$$\therefore I=\int \frac{-dt}{(1-t)(1+t)(1+2t)}$$
$$=\int \frac{dt}{(t-1)(t+1)(1+2t)}$$
$$\text{Let }\frac{1}{(t-1)(t+1)(1+2t)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{1+2t}$$
$$\Rightarrow \frac{1}{(t-1)(t+1)(1+2t)}=\frac{A(t+1)(1+2t)+B(t-1)(1+2t)+C(t-1)(t+1)}{(t-1)(t+1)(1+2t)}$$
$$\Rightarrow 1=A(t+1)(1+2t)+B(t-1)(1+2t)+C(t-1)(t+1)$$
$$\text{Putting t + 1 = 0}$$
$$\Rightarrow t=-1$$
$$1=B(-1-1)(1-2)$$
$$\Rightarrow 1=B(-2)(-1)$$
$$\Rightarrow B=\frac{1}{2}$$
$$\text{Putting t - 1 = 0}$$
$$\Rightarrow t=1$$
$$1=A(1+1)(1+2)$$
$$\Rightarrow 1=A\left(2\right)\left(3\right)$$
$$\Rightarrow A=\frac{1}{6}$$
$$\text{Putting 1 + 2t = 0}$$
$$t=-\frac{1}{2}$$
$$\Rightarrow 1=A\times 0+B\times 0+C(-\frac{1}{2}-1)(-\frac{1}{2}+1)$$
$$1=C(-\frac{3}{2})\left(\frac{1}{2}\right)$$
$$C=-\frac{4}{3}$$
Then,
$$I=\frac{1}{6}\int \frac{dt}{t-1}+\frac{1}{2}\int \frac{dt}{t+1}-\frac{4}{3}\int \frac{dt}{1+2t}$$
$$=\frac{1}{6}\log |t-1|+\frac{1}{2}\log |t+1|-\frac{4}{3}\times \frac{\log |1+2t|}{2}+C$$
$$=\frac{1}{6}\log |t-1|+\frac{1}{2}\log |t+1|-\frac{2}{3}\log |1+2t|+C$$
$$=\frac{1}{6}\log |\cos x-1|+\frac{1}{2}\log |\cos x+1|-\frac{2}{3}\log |1+2\cos x|+C$$