Question

Solve:
$I={\int }_{-\pi /2}^{\pi /2}\sin xdx$

Answer 1

Consider the given integral.

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin xdx$

$I={[-\cos x]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$I=-[\cos \frac{\pi }{2}-\cos (-\frac{\pi }{2})]$

$I=0$

Hence, this is the answer.