Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{\frac{-\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\sin }^{3}xdx \\ ={\int }_{\frac{-\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin x\left(1-{\cos }^{2}x\right)dx \\ ={\int }_{\frac{-\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin xdx-{\int }_{\frac{-\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin x{\cos }^{2}xdx \\ ={\left[-\cos x\right]}_{\frac{-\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}+{\left[\frac{{\cos }^{3}x}{3}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =0+0=0 \end{gathered}$$