Question

The value of $\int \frac{1}{1-\sin x}dx$ is _____.

Answer 1

$I=\int \frac{1}{1-\sin x}dx$

Assume $u=\tan \frac{x}{2}\Rightarrow du=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

Now, $\sin x=\frac{2u}{u^2+1}$

Therefore:

$\int \frac{1}{1-\sin x}dx=\int \frac{2du}{u^2+1-2u}$

$\Rightarrow I=\int \frac{2du}{(u-1{)}^2}$

$\Rightarrow I=\frac{2}{1-u}+C$

$\Rightarrow I=\frac{2\cos \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}+C$