Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}x\sin xdx \\ \mathrm{Integrating}\mathrm{by}\mathrm{parts} \\ I={\left[-x\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\int }_0^{\frac{\mathrm{\pi }}{2}}1\left(-\cos x\right)dx \\ \Rightarrow I={\left[-\mathrm{x}\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[\sin x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=0+1 \\ \Rightarrow I=1 \end{gathered}$$