Answer
\begin{aligned}
\int_{0}^{\pi / 2} \cos^2 x d x&=\frac{\pi}{4}\\
\int_{0}^{\pi / 2} \sin ^2x d x&= \frac{\pi}{4}
\end{aligned}
Work Step by Step
Since we have
$$\int_{0}^{\pi / 2} f(\cos x) d x=\int_{0}^{\pi / 2} f(\sin x) d x$$
Then
$$\int_{0}^{\pi / 2} \cos^2 x d x=\int_{0}^{\pi / 2} \sin ^2x d x$$
Use the rule
$$\cos^2x +\sin^2x=1 $$
We get
\begin{aligned}
\int_{0}^{\pi / 2} \cos^2 x d x+ \int_{0}^{\pi / 2} \sin ^2x d x&=\int_{0}^{\pi / 2} (\cos^2x+\sin ^2x) d x\\
&= \int_{0}^{\pi / 2} dx\\
&= x\bigg|_{0}^{\pi / 2}\\
&=\frac{\pi}{2}
\end{aligned}
Hence
\begin{aligned}
\int_{0}^{\pi / 2} \cos^2 x d x&=\frac{\pi}{4}\\
\int_{0}^{\pi / 2} \sin ^2x d x&= \frac{\pi}{4}
\end{aligned}