Does it hold? $\int_{0}^{\pi/2}\cos^{2k}xdx=\frac{(2k-1)!!}{2k!!}\frac{\pi}{2}$ Does it hold?
$$\int_{0}^{\Large\frac\pi2}\cos^{2k}x\ dx=\frac{(2k-1)!!}{2k!!}\cdot\frac{\pi}{2}$$
where $k \in \mathbb Z$
I have seen some examples like $$\int_{0}^{\Large\frac\pi2}\cos^{6}x\ dx=\frac{5 \cdot 3}{6 \cdot 4 \cdot 2}\cdot\frac{\pi}{2}$$ and I conclude something like this.
I try to use Taylor series but it doesn't work well. Do you have good idea? Thank you a lot!
 A: Let
$$
I_n=\int\cos^n x\ dx.
$$
Using integration by reduction formula, we will obtain
\begin{align}
\int\cos^nx\ dx&=\frac1n\cos^{n-1}x\sin x+\frac{n-1}{n}\int\cos^{n-2}x\ dx\\
I_n&=\frac1n\cos^{n-1}x\sin x+\frac{n-1}{n}I_{n-2}.
\end{align}
It turns out that
$$
\int_0^{\Large\frac\pi2}\cos^nx\ dx=\frac{n-1}{n}\int_0^{\Large\frac\pi2}\cos^{n-2}x\ dx.\tag1
$$
Applying $(1)$ for several times, we will obtain the pattern that leads to the result
$$
\int_0^{\Large\frac\pi2}\cos^nx\ dx=\large\color{green}{\frac{(n-1)!!}{n!!}\cdot\frac\pi2},\tag2
$$
where $I_0=\dfrac\pi2$.
Setting $n=2k$ to $(2)$ yields
$$
\int_0^{\Large\frac\pi2}\cos^{2k}x\ dx=\large\color{blue}{\frac{(2k-1)!!}{(2k)!!}\cdot\frac\pi2}.
$$
A: Another approach:
Consider Beta function
$$
\text{B}(u,v)=2\int_0^{\Large\frac\pi2}(\sin x)^{2u-1}(\cos x)^{2v-1}\ dx=\frac{\Gamma(u)\cdot\Gamma(v)}{\Gamma(u+v)},
$$
then
$$
\int_0^{\Large\frac\pi2}\cos^{2k}x\ dx=\frac{\Gamma\left(\dfrac12\right)\cdot\Gamma\left(k+\dfrac12\right)}{2\cdot\Gamma\left(k+1\right)}=\large\color{blue}{\frac{(2k-1)!!}{(2k)!!}\cdot\frac\pi2},
$$
where


*

*$\Gamma(k)=k!$

*$\Gamma\left(k+\dfrac12\right)=\dfrac{(2k-1)!!}{2^k}\sqrt{\pi}$

*$(2k)!!=2^k\cdot k!$

*$\Gamma\left(\dfrac12\right)=\sqrt\pi$


for $k=0,1,2,3,\cdots$.
The proof of above identities can be seen here and here.
