Explanation

(b)
$\int_{}^{}{e^{x}\left( \ \frac{1 - \sin x}{1 - \cos x}\ \right)\ dx} = \int_{}^{}{e^{x}\left( \frac{1 - \sin x}{2\sin^{2}\frac{x}{2}} \right)dx}$
$= \int_{}^{}{e^{x}\left( \frac{1}{2}cosec^{2}\frac{x}{2} - \cot{\frac{x}{2}\ } \right)dx}$
$\frac{1}{2}\int_{}^{}{e^{x}\ cosec^{2}\frac{x}{2}dx - \int_{}^{}{e^{x}\cot{\frac{x}{2}dx}}}$
$= \frac{1}{2}\int_{}^{}{e^{x}\ cosec^{2}\frac{x}{2}dx - e^{x}\cot{\frac{x}{2}\int_{}^{}{e^{x}cosec^{2}\frac{x}{2}dx\ }}}$
$= e^{x}(\ - \cot{\frac{x}{2}\ ) + c}$
$= - e^{x}\cot{\frac{x}{2} + c}$