Question

Find:
${\int }_0^{\pi }({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})dx$

Answer 1

${\int }_0^{\pi }({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})dx$

$={\int }_0^{\pi }-({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})dx$

$=-{\int }_0^{\pi }\cos 2\left(\frac{x}{2}\right)dx$

$=-{\int }_0^{\pi }\cos xdx$

$=-{\left[\sin x\right]}_0^{\pi }$

$=-(\sin \pi -\sin 0)$

$=0$