Answer
\begin{aligned}
\int_{0}^{\pi / 2} f(\cos x) d x &=\int_{0}^{\pi / 2} f(\sin x) d x
\end{aligned}
Work Step by Step
Given
$$\int_{0}^{\pi / 2} f(\cos x) d x $$
Since
\begin{aligned}
\int_{0}^{\pi / 2} f(\cos x) d x &=\int_{0}^{\pi / 2} f\left(\sin \left(\frac{\pi}{2}-x\right)\right)d x
\end{aligned}
let $ u=\frac{\pi}{2}-x, $then $d u=-d x$ and
\begin{aligned}
\text{at }\ x&= 0\ \to u= \pi/2\\
\text{at }\ x&= \pi/2\ \to u= 0
\end{aligned}
It follows that
\begin{aligned}
\int_{0}^{\pi / 2} f(\cos x) d x &=\int_{0}^{\pi / 2} f\left(\sin \left(\frac{\pi}{2}-x\right)\right)d x \\
&=\int_{\pi / 2}^{0} f(\sin u)(-d u)\\
&=\int_{0}^{\pi / 2} f(\sin u) d u\\
&=\int_{0}^{\pi / 2} f(\sin x) d x
\end{aligned}
