Question

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

Answer 1

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

Put $\sin x=t\Rightarrow \cos xdx=dt$

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

$\therefore \frac{2}{(1-t)(1+t^2)}=\frac{A}{1-t}+\frac{Bt+C}{1+t^2}$
$2=A(1+t^2)+(Bt+C)(1-t)$
$2=A+A{t}^2+Bt+C-B{t}^2-Ct$
$2=(A+C)+(A-B)t^2+(B-C)t$
Comparing both sides, we get,
$A-B=0\Rightarrow A=B\text{and}B-C=0\Rightarrow B=C$
$A+C=2\Rightarrow A+A=2\Rightarrow 2A=2\Rightarrow A=1$
$\therefore A=B=C=1$
$\frac{2}{(1-t)(1+t^2)}=\frac{1}{1-t}+\frac{t+1}{1+t^2}$

$\therefore I=\int \frac{1}{1-t}dt+\int \frac{t+1}{1+t^2}dt$
$=\int \frac{1}{1-t}dt+\int \frac{t}{1+t^2}dt+\int \frac{1}{1+t^2}dt$
$=\int \frac{1}{1-t}dt+\frac{1}{2}\int \frac{2t}{1+t^2}dt+\int \frac{1}{1+t^2}dt$

Let $u=1+t^2$ in the middle integral. $\therefore du=2tdt$

$=\int \frac{1}{1-t}dt+\frac{1}{2}\int \frac{1}{u}du+\int \frac{1}{1+t^2}dt$

= $\frac{\log |1-t|}{-1}+\frac{1}{2}\log |u|+{\tan }^{-1}t+C$

$=-\log |1-\sin x|+\frac{1}{2}\log |1+{\sin }^{2}x|+{\tan }^{-1}\left(\sin x\right)+C$