Question

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

Answer 1

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

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

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

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

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

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

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

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

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

Multiply & divide by $2$

$=-\log (t-1)+\frac{1}{2}\int \frac{2t}{t^2+1}.dt+{\tan }^{-1}\left(t\right)+c$.

$\Downarrow$

$t^2+1=0\Rightarrow 2tdt=du$

$=-\log (t-1)+\frac{1}{2}\int \frac{du}{u}+{\tan }^{-1}\left(t\right)+C$

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

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

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

$\therefore \int \frac{2\cos 0}{(2+\sin x)(1-\sin x)}dx={\tan }^{-1}\left(\sin x\right)+\frac{1}{2}\log ({\sin }^{2}x+1)-\log (\sin x-1)+C$