Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}x{\cos }^{2}xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\mathrm{\pi }}x{\cos }^{2}xdx...\left(\mathrm{i}\right) \\ ={\int }_0^{\mathrm{\pi }}\left(\mathrm{\pi }-\mathrm{x}\right){\cos }^2\left(\mathrm{\pi }-\mathrm{x}\right)dx \\ ={\int }_0^{\mathrm{\pi }}\left(\mathrm{\pi }-\mathrm{x}\right){\cos }^{2}xdx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\mathrm{\pi }}\left(\mathrm{x}+\mathrm{\pi }-\mathrm{x}\right){\cos }^2\mathrm{x}\mathrm{d}\mathrm{x} \\ ={\int }_0^{\mathrm{\pi }}\mathrm{\pi }{\cos }^2\mathrm{x}\mathrm{d}\mathrm{x} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1+\cos 2\mathrm{x}}{2}\mathrm{dx} \\ =\frac{\mathrm{\pi }}{2}{\int }_0^{\mathrm{\pi }}\left(1+\cos 2\mathrm{x}\right)\mathrm{dx} \\ =\frac{\mathrm{\pi }}{2}{\left[\mathrm{x}+\frac{\sin 2\mathrm{x}}{2}\right]}_0^{\mathrm{\pi }} \\ =\frac{\mathrm{\pi }}{2}\left(\mathrm{\pi }-0\right) \\ \mathrm{Hence}I=\frac{{\mathrm{\pi }}^2}{4} \end{gathered}$$