Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}\left({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\mathrm{\pi }}\left({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2}\right)\mathit{}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I=-{\int }_0^{\mathrm{\pi }}\cos xdx\left[\because \cos A={\cos }^2\frac{A}{2}-{\sin }^2\frac{A}{2}\right] \\ \Rightarrow I=-{\left[\sin x\right]}_0^{\mathrm{\pi }} \\ \Rightarrow I=0 \end{gathered}$$