Integral of $\int_0^{2\pi}\cos^n(x)\,dx$.

Consider

$$\int_0^{2\pi}\cos^n(x)\,dx,\qquad n\text{ a positive integer}$$ For $n$ odd, the answer is zero.

Is there a slick way to find a closed form for $n$ even?

Funny enough, someone just posted a question on the Power-reduction formula two hours ago. Using that, you readily get the result

$$\frac{2\pi}{2^n}\binom{n}{n/2}\;.$$

Qiaochu Yuan's hint seems to be the simplest approach: By the binomial theorem for any $n\geq0$ one has $$2^n\cos^n x=(e^{ix}+e^{-ix})^n=\sum_{k=0}^n {n\choose k} (e^{ix})^k\ (e^{-ix})^{n-k}=\sum_{k=0}^n {n\choose k} e^{(2k-n)ix}\ .\qquad(*)$$ Since $$\int_0^{2\pi}e^{i\ell x}\ dx=\cases{2\pi&$\quad(\ell=0)$\cr 0&$\quad(\ell\ne0)$\cr}$$ at most one term on the right of $(*)$ contributes to the integral $J_n:=\int_0^{2\pi}\cos^n x\ dx$. When $n$ is odd then $2k-n\ne0$ for all $k$ in $(*)$, therefore $J_n=0$ in this case. When $n$ is even then $k=n/2$ gives the only contribution to the integral, and we get $$\int_0^{2\pi} \cos^n x\ dx={2\pi\over 2^n}{n\choose n/2}\ .$$

It is also possible with partial integration, though getting the closed formula from the other solution is not as easy to see.

$$ C(n):=\int_0^{2\pi}\!\!\!\cos^n(x)\,dx =\int_0^{2\pi}\!\!\!\cos^{n-1}(x)\cos(x)\,dx $$ partial integration gives

$$ = (n-1)\int_0^{2\pi}\!\!\!\cos^{n-2}(x)\sin^2(x)\,dx$$ $$ =(n-1)\int_0^{2\pi}\!\!\!\cos^{n-2}(x)\left(1-\cos^2(x)\right)\,dx $$ $$ \Rightarrow \int_0^{2\pi}\!\!\!\cos^n(x)\,dx = \frac{n-1}{n}\int_0^{2\pi}\!\!\!\cos^{n-2}(x)\,dx $$

So in short: $C(0)=2\pi$, $C(1)=0$ and $$C(n)=\frac{n-1}{n}C(n-2) = \frac{(n-1)!!}{n!!} 2\pi\quad \text{for }n\text{ even} .$$

It's possible to do this integral in a couples of lines using the residue theorem from complex analysis.

Details: The usual trick to do definite integrals going from $0$ to $2\pi$ is to let $\cos x = \dfrac {z^2 + 1} {2z}$ where $z = {\rm e} ^{{\rm i} x}$. This substitution also implies that ${\rm d} x = \dfrac {{\rm d} z} {{\rm i} z}$. Then this is reduced to the contour integral of $\left( \dfrac {z^2 + 1} {2z} \right) ^n \dfrac {{\rm d} z} {{\rm i} z}$ where the contour is the unit circle in the complex plane. Then you can expand this using the binomial theorem and get the coefficient of $\dfrac 1 z$ and apply the residue theorem to get the answer.

Let $X$ be a standard normal random variable. Then, your integral $I$ can be computed as

$$ I = 2\pi\cdot \mathbb E[X^n] = \begin{cases}2\pi\cdot (n-1)!! = \frac{2\pi}{2^n}{n\choose n/2},&\mbox{ if }n\text{ is even},\\ 0,&\mbox{ else.} \end{cases} $$