Question

$\text{Find}:\int \frac{\cos x}{(1+\sin x)(2+\sin x)}dx$

Answer 1

$\text{Let}I=\int \frac{\cos x}{(1+\sin x)(2+\sin x)}dx\text{Put}\sin x=t\Rightarrow \cos xdx=dt\Rightarrow I=\int \frac{dt}{(1+t)(2+t)}\text{Now, let}\frac{1}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}\Rightarrow \frac{1}{(1+t)(2+t)}=\frac{A(2+t)+B(1+t)}{(1+t)(2+t)}=\frac{2A+B+(A+B)t}{(1+t)(2+t)}$

Comparing the numerator, we get
$A+B=0\cdots \left(1\right)2A+B=1\cdots \left(2\right)$
Solving $\left(1\right)\&\left(2\right),$ we get
$A=1,B=-1\therefore I=\int \frac{1}{1+t}dt-\frac{1}{2+t}dt\Rightarrow I=\log |1+t|-\log |2+t|+C\Rightarrow I=\log \left|\frac{1+t}{2+t}\right|+C\Rightarrow I=\log \left|\frac{1+\sin x}{2+\sin x}\right|+C$