Question

Calculate the following integrals.
${\int }_0^{\pi /2}sinxcosxdx.$

Answer 1

${\int }_0^{\pi /2}\sin x\cos xdx$

$=\frac{1}{2}{\int }_0^{\pi /2}2\sin x\cos xdx$

$=\frac{1}{2}{\int }_0^{\pi /2}\sin 2xdx$

$=\frac{1}{2}{\left[\frac{-\cos 2x}{2}\right]}_0^{\pi /2}$

$=\frac{-1}{4}{\left[\cos 2x\right]}_0^{\pi /2}$

$=\frac{1}{4}$