Question

$\underset{\mathrm{\pi }/2}{\overset{\mathrm{\pi }}{\int }}{e}^x\left(\frac{1-\sin x}{1-\cos x}\right)dx$

Answer 1

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