Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}{\sin }^{3}xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^{3}x\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin x{\sin }^{2}xdx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin x\left(1-{\cos }^{2}x\right)dx \\ \mathrm{Let}u=\cos x,du=-\sin xdx \\ \therefore I=\int -\left(1-u^2\right)du \\ \Rightarrow I=\left[\frac{u^3}{3}-u\right] \\ \Rightarrow I={\left[\frac{{\cos }^{3}x}{3}-\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=0-\frac{1}{3}+1 \\ \Rightarrow I=\frac{2}{3} \end{gathered}$$