Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}{x}^2{\cos }^{2}x\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}{x}^2\left(\frac{1+\cos 2x}{2}\right)dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\frac{x^2}{2}+\frac{x^2\cos 2x}{2}\right)dx \\ \Rightarrow I={\left[\frac{x^3}{6}\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[\frac{x^2\sin 2x}{4}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{x}{2}\sin 2x\mathit{}\mathit{d}x \\ \Rightarrow I={\left[\frac{x^3}{6}\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[\frac{x^2\sin 2x}{4}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\left[\frac{-x\cos 2x}{4}\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\int }_0^{\frac{\mathrm{\pi }}{2}}-1\frac{\cos 2x}{2}dx \\ \Rightarrow I={\left[\frac{x^3}{6}\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[\frac{x^2\sin 2x}{4}\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[\frac{x\cos 2x}{4}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\left[\frac{\sin 2x}{4}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=\frac{{\mathrm{\pi }}^3}{48}-\frac{\mathrm{\pi }}{8} \end{gathered}$$