Evaluate $\int_{0}^{\pi/2}\cos^{2n+1}(x)dx$ How can I compute this Integral for integer $n$ from $0$ to $\pi/2$
$$\int_{0}^{\pi/2}\cos^{2n+1}(x)dx?$$

Keywords: Wallis integrals.
For $k\geq 0$ integer, define
\begin{align}J_k=\int_0^{\frac{\pi}{2}}\cos^{2k+1} x\,dx\end{align}
Perform integration by parts, ($(\sin  x)^\prime=\cos x$ )
\begin{align}J_{k+1}&= \int_0^{\frac{\pi}{2}}\cos^{2k+3} x\,dx\\
&=\Big[\sin x\cos^{2k+2} x\Big]_0^{\frac{\pi}{2}}+(2k+2)\int_0^{\frac{\pi}{2}}\sin^ 2x\cos^{2k+1} x\,dx\\
&=(2k+2)\int_0^{\frac{\pi}{2}}\sin^ 2x\cos^{2k+1} x\,dx\\
&=(2k+2)\int_0^{\frac{\pi}{2}}\left(1-\cos^ 2x\right)\cos^{2k+1} x\,dx\\
&=(2k+2)\left(J_{k}-J_{k+1}\right)
\end{align}
Therefore,
\begin{align}J_{k+1}&=\frac{2k+2}{2k+3}J_k\end{align}
Observe that,
\begin{align}J_0&=\int_0^{\frac{\pi}{2}} \cos x\,dx\\
&=\Big[\sin x\Big]_0^{\frac{\pi}{2}}\\
&=1
\end{align}
Addendum:
One obtains:
\begin{align}\boxed{J_k=\frac{2^kk!}{\big(2(k-1)+3\big)\big(2(k-2)+3\big)...\times 3}}\end{align}
