Question

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

Answer 1

We have,
$$I=\int \frac{\sin 2xdx}{(1+\sin x)(2+\sin x)}$$
$$=\int \frac{2\sin x\cos xdx}{(1+\sin x)(2+\sin x)}$$
Putting sin x = t

$$\Rightarrow \cos xdx=dt$$
$$\therefore I=\int \frac{2tdt}{(1+t)(2+t)}$$
$$=2\int \frac{tdt}{(1+t)(2+t)}$$
$$\text{Let }\frac{t}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}$$
$$\Rightarrow \frac{t}{(1+t)(2+t)}=\frac{A(2+t)+B(1+t)}{(1+t)(2+t)}$$
$$\Rightarrow t=A(2+t)+B(1+t)$$
Putting 2 + t = 0

$$\Rightarrow t=-2$$
$$-2=A\times 0+B(-2+1)$$
$$\Rightarrow -2=B(-1)$$
$$\Rightarrow B=2$$
$$\text{Let }t+1=0$$
$$t=-1$$
$$\Rightarrow -1=A(-1+2)+B\times 0$$
$$A=-1$$
$$\therefore I=2\int (\frac{-1}{t+1}+\frac{2}{t+2})dt$$
$$=2[-\log |t+1|+2\log |t+2|]+C$$
$$=4\log |t+2|-2\log |t+1|+C$$
$$=\log \left|\frac{{(t+2)}^4}{{(t+1)}^2}\right|+C$$
$$=\log \left|\frac{{(\sin x+2)}^4}{{(\sin x+1)}^2}\right|+C$$