Question

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

Answer 1

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