Question

$\underset{-\mathrm{\pi }/4}{\overset{\mathrm{\pi }/4}{\int }}{\sin }^{2}xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\sin }^{2}xdx \\ \mathrm{Here}f\left(x\right)={\sin }^{2}x \\ f\left(-x\right)={\sin }^2\left(-x\right)={\sin }^{2}x=f\left(x\right) \\ \mathrm{Hence}{\sin }^{2}xi\mathrm{s}\mathrm{an}\mathrm{even}\mathrm{function} \\ \mathrm{Therefore}, \\ I=2{\int }_0^{\frac{\mathrm{\pi }}{4}}{\sin }^2\mathrm{x}d\mathrm{x} \\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}\left(\frac{1-\cos 2\mathrm{x}}{2}\right)\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\left(1-\cos 2\mathrm{x}\right)\mathrm{dx} \\ ={\left[\mathrm{x}-\frac{\sin 2\mathrm{x}}{2}\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ =\frac{\mathrm{\pi }}{4}-\frac{1}{2} \end{gathered}$$→