Answer
$\int_{0}^{\frac{\pi}{2}} \cos ^{n} x d x=\int_{0}^{\frac{\pi}{2}} \sin ^{n} x d x$
Work Step by Step
We need to prove : ${\int_{0}^{\frac{\pi}{2}} \cos ^{n} x d x=\int_{0}^{\frac{\pi}{2}} \sin ^{n} x d x}$
$\int_{0}^{\frac{\pi}{2}} \cos ^{n}\left(-x+\frac{\pi}{2}\right) d x =\int_{0}^{\frac{\pi}{2}} \sin ^{n} x d x \quad \because \cos \left(-x+\frac{\pi}{2}=\sin x\right)$
$\operatorname{Let} -x+\frac{\pi}{2}=u \Rightarrow d u=-d x$
When $\frac{\pi}{2}=x \Rightarrow u=0$
When $x=0 \Rightarrow u=\frac{\pi}{2}$
\[
\begin{aligned}
\therefore \int_{0}^{\frac{\pi}{2}} \cos ^{n}\left(-x+\frac{\pi}{2}\right) d x &=-\int_{\frac{\pi}{2}}^{0} \cos ^{n} u d u \\
&=\int_{0}^{\frac{\pi}{2}} \cos ^{n} u d u
\end{aligned}
\]
Substitute $x=u \Rightarrow d x=d u$
\[
\therefore \int_{0}^{\frac{\pi}{2}} \cos ^{n} x d x=\int_{0}^{\frac{\pi}{2}} \sin ^{n} x d x
\]
