Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}{x}^2\cos 2x\mathit{d}x.\mathrm{Then}, \\ \mathrm{Integrating}\mathrm{by}\mathrm{parts} \\ I={\left[x^2\mathit{}\frac{\sin 2x}{2}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\int }_0^{\frac{\mathrm{\pi }}{2}}2x\mathit{}\frac{\sin 2x}{2}\mathit{d}x \\ \Rightarrow I={\left[x^2\mathit{}\frac{\sin 2x}{2}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\left[-x\frac{\cos 2x}{2}\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\int }_0^{\frac{\mathrm{\pi }}{2}}-1\frac{\cos 2x}{2}\mathit{}\mathit{d}x \\ \Rightarrow I={\left[x^2\mathit{}\frac{\sin 2x}{2}\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[x\mathit{}\frac{\cos 2x}{2}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\left[\frac{\sin 2x}{4}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=0-\frac{\mathrm{\pi }}{4}-0 \\ \Rightarrow I=-\frac{\mathrm{\pi }}{4} \end{gathered}$$