$\int\limits_{\pi /2}^\pi {{e^x}\left( {\cfrac{{1 - \sin x}}{{1 - \cos x}}} \right)} dx$

Let $I = \int\limits_{\pi /2}^\pi {{e^x}\left( {\cfrac{{1 - \sin x}}{{1 - \cos x}}} \right)} dx = \int\limits_{\pi /2}^\pi {{e^x}\left( {\cfrac{{1 - 2\sin \cfrac{x}{2}\cos \cfrac{x}{2}}}{{2{{\sin }^2}\cfrac{x}{2}}}} \right)} dx$

$= \int\limits_{\pi /2}^\pi {{e^x}\left( {\cfrac{1}{2}\cos e{c^2}\cfrac{x}{2} - \cot \cfrac{x}{2}} \right)dx = - \int\limits_{\pi /2}^\pi {{e^x}\left( {\cot \cfrac{x}{2} - \cfrac{1}{2}\cos e{c^2}\cfrac{x}{2}} \right)dx} }$

$= - \left[ {{e^x}\cot \cfrac{x}{2}} \right]_{\pi /2}^\pi = - \left[ {{e^\pi }\cot \cfrac{\pi }{2} - {e^{\pi /2}}\cot \cfrac{\pi }{4}} \right] = - \left[ {0 - {e^{\pi /2}} \cdot 1} \right] = {e^{\pi /2}}$