Question

Solve:
${\int }_{-\pi /2}^{\pi /2}\cos xdx$.

Answer 1

$I={\int }_{-\pi /2}^{\pi /2}\cos xdx$
$=\sin x{|}_{-\pi /2}^{\pi /2}$
$=\sin \frac{\pi }{2}-\sin (-\frac{\pi }{2})$
$=1+1=3[\because {\int }_a^{b}f(x)dx=I(x){|}_a^b=I(b)-I(a\left)\right]$.