Question

Evaluate

${\int }_{\frac{}{}}^{\frac{}{}}\frac{1}{1-\sin x}dx$

Answer 1

$I={\int }^{}\frac{1}{1-\sin \left(x\right)}dx$

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

$I=-{\int }^{}\frac{{\sec }^2\frac{x}{2}}{{\left(\tan \frac{x}{2}-1\right)}^2}dx$

Substitute

$u=\tan \frac{x}{2}=-2{\int }^{}\frac{1}{u^2}du$

$I=-\frac{2}{\tan \frac{x}{2}-1}+C$

Hence, this is the answer.