Showing that $\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$

Question

As the title suggests, I would like to show that: $$\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$$ for every positive integer $n$.

How can I go about doing that? I understand the that there are already solutions for the problem: $\frac{1}{2\pi}\int^{2\pi}_{0}(\cos\theta)^{2n}d\theta= {2n \choose n} \frac{\pi}{2^{2n-1}}$. Is there any modifications I can do to obtain the desired solution?

Answer 1

Your answer for $\int_0^{2\pi} (\cos \theta)^{2n}d\theta$ is false (it is one of Wallis' integrals).

$$ W_{2p}=\int_0^{\frac{\pi}{2}} (\cos \theta)^{2p}d\theta =\frac\pi2\prod_{k=1}^p\frac{2k-1}{2k}=\frac\pi2\frac{(2p)!}{(2^pp!)^2} $$

Changing the boundaries is easily feasible considering the periodicity and parity of $x\mapsto cos^{2n}(x)$. Thus:

$$\frac{1}{2\pi}\int_0^{2\pi} (\cos \theta)^{2n}d\theta = \frac{2}{\pi}W_{2n}$$

Then answering your question is quite straightforward: $$ \frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{2^{2n+1}}{\pi}W_{2n}=\frac{(2n)!}{(n!)^2} $$

Answer 2

Integrating by parts

\begin{align*} \int_0^{2\pi}\cos^{2n}\theta d\theta&=\Big[\cos^{2n-1}\theta\sin\theta\Big]_0^{2\pi}-\int_0^{2\pi}\sin\theta(2n-1)\cos^{2n-2}\theta(-\sin\theta) d\theta\\ &=(2n-1)\int_0^{2\pi}(\cos^{2n-2}\theta-\cos^{2n}\theta)d\theta\\ &=(2n-1)\int_0^{2\pi}(\cos^{2n-2}\theta-\cos^{2n}\theta)d\theta\\ \end{align*}

So, we have

\begin{align*} \int_0^{2\pi}\cos^{2n}\theta d\theta&=\frac{2n-1}{2n}\int_0^{2\pi}\cos^{2(n-1)}\theta d\theta\\ &=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\int_0^{2\pi}\cos^{2(n-2)}\theta d\theta\\ &=\dots\\ &=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdots\frac{1}{2}\cdot\int_0^{2\pi}d\theta\\ &=\left(\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdots\frac{1}{2}\right)\left(\frac{2n}{2n}\cdot\frac{2n-2}{2n-2}\cdot\frac{2n-4}{2n-4}\cdots\frac{2}{2}\right)\left(2\pi\right)\\ &=\;\frac{2\pi(2n)!}{2^{2n}(n!)^2} \end{align*}

Answer 3

You can solve the integral by induction. First of all, you may notice that

\begin{align*} \frac{1}{2\pi}\int_0^{2\pi}(2\cos\theta)^2 d\theta&=\frac{1}{\pi}\Big[\theta+\cos\theta\sin\theta\Big]_0^{2\pi}=2=\frac{2!}{1!1!} \end{align*}

Thus the case $n=1$ holds. Now, integrating by parts:

\begin{align*} \frac{1}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n} d\theta&=\frac{2^{2n}}{2\pi}\Big(\Big[\cos^{2n-1}\theta\sin\theta\Big]_0^{2\pi}+(2n-1)\int_0^{2\pi}\sin^2\theta\cos^{2n-2}\theta d\theta\Big)\\ &=\frac{2^{2n}(2n-1)}{2\pi}\int_0^{2\pi}(\cos^{2n-2}\theta-\cos^{2n}\theta)d\theta\\ \end{align*}

then

\begin{align*} \frac{1}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n} d\theta+\frac{(2n-1)}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n}d\theta&=\\ \frac{2n}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n}d\theta&=\\ &=\frac{2^{2n}(2n-1)}{2\pi\cdot2^{2n-2}}\int_0^{2\pi}(2\cos\theta)^{2n-2}d\theta\\ \end{align*}

that is

\begin{align*} \int_0^{2\pi}(2\cos\theta)^{2n} d\theta &=\frac{2\pi}{2n}\cdot\frac{2^{2n}(2n-1)}{2\pi\cdot2^{2n-2}}\int_0^{2\pi}(2\cos\theta)^{2n-2} d\theta=\\ &=\frac{2^2(2n-1)}{2n}\cdot\frac{(2n-2)!}{(n-1)!(n-1)!}=\frac{2n(2n-1)!}{n!n!}=\frac{(2n)!}{n!n!} \end{align*}