Integral of $\cos^{n}(x)$ on $$ for $n$ a positive integer

$$\int_0^{2\pi}\cos^n(x)\,dx,\qquad n\text{ a positive integer}$$

For $n$ odd, the answer is zero. Is there a slick way to find a closed form for $n$ even?

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Qiaochu Yuan’s hint seems to be the simplest approach: By the binomial theorem for any $n\geq0$ one has
$$2^n\cos^n x=(e^{ix}+e^{-ix})^n=\sum_{k=0}^n {n\choose k} (e^{ix})^k\ (e^{-ix})^{n-k}=\sum_{k=0}^n {n\choose k} e^{(2k-n)ix}\ .\qquad(*)$$
$$\int_0^{2\pi}e^{i\ell x}\ dx=\cases{2\pi&$\quad(\ell=0)$\cr 0&$\quad(\ell\ne0)$\cr}$$
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
$$\int_0^{2\pi} \cos^n x\ dx={2\pi\over 2^n}{n\choose n/2}\ .$$

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


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.