Question

Calculate the following integrals.
${\int }_{-\pi /2}^{\pi /2}|sinx|dx.$

Answer 1

${\int }_{-\pi /2}^{\pi /2}|\sin x|dx$

We know that $|\sin (-x)|=|sinx|$ is an even function

$=2{\int }_0^{\pi /2}|\sin x|dx$

And, $|\sin x|=\sin x$ in the interval $(0,\pi /2)$

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

$=2[-\cos x{]}_0^{\pi /2}$

$=2$