Question

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

Answer 1

\[\text{Let }\ I = \int_\frac{\pi}{2}^\pi e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) d x . Then, \]
\[I = \int_\frac{\pi}{2}^\pi e^x \left( \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx .................\left[ As, \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}, \cos A = 1 - 2 \sin^2 \frac{A}{2} \right]\]
\[ \Rightarrow I = \int_\frac{\pi}{2}^\pi e^x \left( \frac{1}{2} {cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx\]
\[ \Rightarrow I = \int_\frac{\pi}{2}^\pi \frac{1}{2} e^x {cosec}^2 \frac{x}{2} dx - \int_\frac{\pi}{2}^\pi e^x \cot \frac{x}{2} dx\]
\[\text{Integrating second term by parts}\]
\[I = \left\{ - \left[ e^x \cot \frac{x}{2} \right]_\frac{\pi}{2}^\pi - \int_\frac{\pi}{2}^\pi \frac{1}{2} e^x {cosec}^2 \frac{x}{2} dx \right\} + \int_\frac{\pi}{2}^\pi \frac{1}{2} e^x {cosec}^2 \frac{x}{2} dx\]
\[ \Rightarrow I = - \left[ 0 - e^\frac{\pi}{2} \right]\]
\[ \Rightarrow I = e^\frac{\pi}{2} \]