Question

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

Answer 1

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

Substituting $\sin x=u\Rightarrow du=\cos xdx$, we get,

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

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

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

Substituting back, we get:

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

where $C$ is a real constant of integration.