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?

• Write $\cos(x) = (e^{ix} + e^{-ix})/2$. I am reasonably certain this is a duplicate but I don't know how I'd go about finding the duplicate. For a combinatorial proof see this post: qchu.wordpress.com/2010/03/07/… Mar 28 '12 at 20:33
• Should be proportional to $2^{-n}|P_n(0)|$ (or so), where $P_n$ is the $n$th Legendre polynomial! Sep 7 '21 at 16:20

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

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

• Fascinating connection! Mar 28 '12 at 20:18

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 $$C(n)=\frac{n-1}{n}C(n-2) = \frac{(n-1)!!}{n!!} 2\pi\quad \text{for }n\text{ even} .$$

• You got a factor of $2$ wrong; it's $nC(n)=(n-1)C(n-2)$ and thus $C(n)=(n-1)/n C(n)$, and that agrees with my answer. Mar 28 '12 at 20:42
• @joriki thank you. already fixed it. Took me a while to see the 2 or 3 mistakes I did (the factor of two was the last one ;) ). Mar 28 '12 at 20:45

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.

• Without more details, this should be a comment and not an answer. Jul 12 '12 at 7:24
• Sorry, here are some details. The usual trick to do definite integrals going from 0 to 2pi is to let cos(x) = (z^2 + 1)/2z where z = e^ix. This substitution also implies that dx = dz/iz. Then this is reduced to the contour integral of ((z^2 + 1)/2z)^n*dz/iz 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 1/z and apply the residue theorem to get the answer. Jul 13 '12 at 8:53
• You should edit your answer instead of posting that as a comment. Jul 13 '12 at 8:56

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

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