Question

${\int }_0^{\frac{\pi }{2}}{x}^{2}co{s}^{2}xdx$

Answer 1

${\int }_0^{\frac{\pi }{2}}{x}^2{\cos }^{2}xdx$

We have $\cos 2x=2{\cos }^{2}x-1\Rightarrow {\cos }^{2}x=\frac{1+\cos 2x}{2}$

$={\int }_0^{\frac{\pi }{2}}{x}^2\left(\frac{1+\cos 2x}{2}\right)dx$

$={\int }_0^{\frac{\pi }{2}}{x}^{2}dx+{\int }_0^{\frac{\pi }{2}}{x}^2\left(\frac{\cos 2x}{2}\right)dx$

$=\frac{1}{2}{\left[\frac{x^3}{3}\right]}_0^{\frac{\pi }{2}}+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{x}^2\cos 2xdx$

$=\frac{1}{2}[\frac{{\pi }^3}{3\times 8}-0]+\frac{1}{2}{I}_1$

where $I_1={\int }_0^{\frac{\pi }{2}}{x}^2\cos 2xdx$

Take $u=x^2\Rightarrow du=2xdx$ and $dv=\cos 2xdx\Rightarrow v=\frac{\sin 2x}{2}$

$\Rightarrow I_1={\int }_0^{\frac{\pi }{2}}{x}^2\cos 2xdx$

$={\left[\frac{x^2\sin 2x}{2}\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\frac{\sin 2x}{2}2x.dx$

$=\frac{1}{2}{\left[x^2\sin 2x\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}x\sin 2xdx$

$=\frac{1}{2}\left[\frac{{\pi }^2}{4}\sin \frac{2\pi }{2}\right]-{\int }_0^{\frac{\pi }{2}}x\sin 2xdx$

$=0-{\int }_0^{\frac{\pi }{2}}x\sin 2xdx$

$=-{\int }_0^{\frac{\pi }{2}}x\sin 2xdx$

Take $u=x\Rightarrow du=dx$ and $dv=\sin 2xdx\Rightarrow v=\frac{-\cos 2x}{2}$

$=-\{{\left[\frac{-x\cos 2x}{2}\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\frac{-\cos 2x}{2}dx\}$

$=-\frac{\frac{\pi }{2}\cos \pi }{2}+0$

$=\frac{\pi }{4}$

$\therefore {\int }_0^{\frac{\pi }{2}}{x}^2{\cos }^{2}xdx$

$=\frac{1}{2}[\frac{{\pi }^3}{3\times 8}-0]+\frac{1}{2}{I}_1$

$=\frac{{\pi }^3}{48}+\frac{1}{2}\times \frac{\pi }{4}$

$=\frac{{\pi }^3}{48}+\frac{\pi }{8}$