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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?

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    $\begingroup$ 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/… $\endgroup$ – Qiaochu Yuan Mar 28 '12 at 20:33
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Qiaochu Yuan's hint seems to be the simplest approach: By the binomial theorem for any $n\geq0$ one has $$2^n\cos^n x=(e^{ix}+e^{-ix})^n=\sum_{k=0}^n {n\choose k} (e^{ix})^k\ (e^{-ix})^{n-k}=\sum_{k=0}^n {n\choose k} e^{(2k-n)ix}\ .\qquad(*)$$ 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 $$\int_0^{2\pi} \cos^n x\ dx={2\pi\over 2^n}{n\choose n/2}\ .$$

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

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

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    $\begingroup$ Fascinating connection! $\endgroup$ – Kirthi Raman Mar 28 '12 at 20:18
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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 $$C(n)=\frac{n-1}{n}C(n-2) = \frac{(n-1)!!}{n!!} 2\pi\quad \text{for }n\text{ even} .$$

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  • $\begingroup$ 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. $\endgroup$ – joriki Mar 28 '12 at 20:42
  • $\begingroup$ @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 ;) ). $\endgroup$ – example Mar 28 '12 at 20:45
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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.

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    $\begingroup$ Without more details, this should be a comment and not an answer. $\endgroup$ – J. M. is a poor mathematician Jul 12 '12 at 7:24
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    $\begingroup$ 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. $\endgroup$ – Ahsan Jul 13 '12 at 8:53
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    $\begingroup$ You should edit your answer instead of posting that as a comment. $\endgroup$ – J. M. is a poor mathematician Jul 13 '12 at 8:56

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