if $f\left(x\right)$ is an odd function of $x,$ then ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}} f(\cos x)dx$ is equal to

$f(\cos x)$ is an even function.

$\begin{array}{l}\because f(\cos (-x\left)\right)=f(\cos x)\\ \therefore {\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}} f(\cos x)dx=2{\int }_0^{\frac{\pi }{2}} f(\cos x)dx=2{\int }_0^{\frac{\pi }{2}} f(\sin x)dx.\end{array}$