Question

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

Answer 1

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