${\mathrm{I}}_2=\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{{\mathrm{x}}^2\mathrm{sinxcos}\left(\frac{\mathrm{\pi }}{2}\mathrm{cosx}\right)\mathrm{dx}}{(2\mathrm{x}-\mathrm{\pi })}$

${\mathrm{I}}_2=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}[\frac{{(\frac{\mathrm{\pi }}{2}-\mathrm{x})}^2\mathrm{cosxcos}\left(\frac{\mathrm{\pi }}{2}\mathrm{sinx}\right)}{-2\mathrm{x}}+\frac{{(\frac{\mathrm{\pi }}{2}+\mathrm{x})}^2\mathrm{cosxcos}\left(\frac{\mathrm{\pi }}{2}\mathrm{sinx}\right)}{2\mathrm{x}}]\mathrm{dx}$]

$=2\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{\mathrm{\pi }}{2}\mathrm{cosxcos}\left(\frac{\mathrm{\pi }}{2}\mathrm{sinx}\right)\mathrm{dx}$

$=2\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\mathrm{cos\theta }\cdot \mathrm{d\theta }=2[\mathrm{Let}\frac{\mathrm{\pi }}{2}\sin \mathrm{x}=\mathrm{\theta }]$

${\mathrm{I}}_1=\underset{-\mathrm{\pi }/2}{\overset{\mathrm{\pi }/2}{\int }}\frac{{\cos }^2\mathrm{xdx}}{1+3^{\mathrm{x}}}$

${\mathrm{I}}_1=\frac{1}{2}\underset{-\mathrm{\pi }/2}{\overset{\mathrm{\pi }/2}{\int }}{\cos }^2\mathrm{x}\mathrm{dx}=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}{\cos }^2\mathrm{x}\mathrm{dx}$

${\mathrm{I}}_1=\frac{\mathrm{\pi }}{4}$