Question

Integrate:

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

Answer 1

Consider the given integral.

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

Let $t=\sin x$

$dt=\cos xdx$

Therefore,

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

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

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

$I=-\ln (t-1)+\frac{1}{2}\ln (t^2+1)+{\tan }^{-1}t+C$

On putting the value of $t$, we get

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

Hence, this is the answer.