$$\displaystyle\int{ \frac{ 1 }{ 1- \sin x } }d x$$
$-\frac{2}{\tan(\frac{x}{2})-1}+С$
$\frac{2\frac{\mathrm{d}}{\mathrm{d}x}(\tan(\frac{x}{2}))}{\left(\tan(\frac{x}{2})-1\right)^{2}}$
