$\mathrm{Evaluate} \mathrm{the} \mathrm{integrals} {\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} {\sin }^2 \mathrm{xdx}$

OR

$\mathrm{Evaluate} \mathrm{the} \mathrm{integrals} {\int }_0^{2\mathrm{\pi }} \frac{1}{1+{\mathrm{e}}^{\sin \mathrm{x}}}\mathrm{dx}$

$\mathrm{Let} \mathrm{I}={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} {\sin }^2\mathrm{xdx}$
$\mathrm{Here}, \mathrm{f}\left(\mathrm{x}\right)={\sin }^2 \mathrm{x}$
$\mathrm{Now}, \mathrm{f}(-\mathrm{x})={\sin }^2 (-\mathrm{x})=[\sin (-\mathrm{x}){]}^2$
${\text{=(−sin x)}}^2[\because \sin (-\mathrm{\theta })=-\sin \mathrm{\theta }]$
$={\sin }^2 \mathrm{x}=\mathrm{f}\left(\mathrm{x}\right)$
$\text{ So, }\mathrm{f}\left(\mathrm{x}\right)\text{ is an even function. }$
$\therefore \mathrm{I}={\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} {\sin }^2 \mathrm{xdx}=2{\int }_0^{\mathrm{\pi }/2} {\sin }^2 \mathrm{xdx}$
$$\begin{gathered} \left[\because {\int }_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=2{\int }_0^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}, \mathrm{if} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{an} \mathrm{even} \mathrm{function}, \\ \mathrm{here} {\sin }^2\mathrm{x} \mathrm{is} \mathrm{an} \mathrm{even} \mathrm{function}.\right] \end{gathered}$$
$=2{\int }_0^{\mathrm{\pi }/2} \left[\frac{1-\cos 2\mathrm{x}}{2}\right]\mathrm{dx}\left[\because \cos 2\mathrm{\theta }=1-2{\sin }^2 \mathrm{\theta }\right]$
$\begin{array}{l}={\int }_0^{\mathrm{\pi }/2} (1-\cos 2\mathrm{x})\mathrm{dx}\\ ={\left[\mathrm{x}-\frac{\sin 2\mathrm{x}}{2}\right]}_0^{\mathrm{\pi }/2}\\ =\left[\frac{\mathrm{\pi }}{2}-\frac{\sin \mathrm{\pi }}{2}\right]-[0-0]=\frac{\mathrm{\pi }}{2}-0=\frac{\mathrm{\pi }}{2}\end{array}$

OR

$\mathrm{Let} \mathrm{I}={\int }_0^{2\mathrm{\pi }} \frac{1}{1+{\mathrm{e}}^{\sin \mathrm{x}}}\mathrm{dx}\to \left(1\right)$
$\therefore \mathrm{I}={\int }_0^{\mathrm{\pi }} \left[\frac{1}{1+{\mathrm{e}}^{\sin \mathrm{x}}}+\frac{1}{1+{\mathrm{e}}^{\sin (2\mathrm{\pi }-x)}}\right]\mathrm{dx}$$\left[\because {\int }_0^{2\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{2a} \left\{\mathrm{f}\right(\mathrm{x})+\mathrm{f}(2\mathrm{a}-\mathrm{x}\left)\right\}\mathrm{dx}\right]$

$={\int }_0^{\mathrm{\pi }} \left[\frac{1}{1+{\mathrm{e}}^{\sin \mathrm{x}}}+\frac{1}{1+{\mathrm{e}}^{-\sin \mathrm{x}}}\right]\mathrm{dx}$
$[\because \sin (2\mathrm{\pi }-\mathrm{\theta })=-\sin \mathrm{\theta }]$
$={\int }_0^{\mathrm{\pi }} \left[\frac{1}{1+{\mathrm{e}}^{\sin \mathrm{x}}}+\frac{{\mathrm{e}}^{\sin \mathrm{x}}}{{\mathrm{e}}^{\sin \mathrm{x}}+1}\right]\mathrm{dx}$
$={\int }_0^{\mathrm{\pi }} \frac{1+{\mathrm{e}}^{\sin \mathrm{x}}}{1+{\mathrm{e}}^{\sin \mathrm{x}}}\mathrm{dx}={\int }_0^{\mathrm{\pi }} 1\mathrm{dx}=[\mathrm{x}{]}_0^{\mathrm{\pi }}=[\mathrm{\pi }-0]=\mathrm{\pi }$