integration of $\int_0^{2\pi} cos^{2n}(t)dt$ Show that for any $n \in \mathbb{N}$, 
$$\frac{1}{2\pi}\int_0^{2\pi}\cos^{2n}(t)dt = \frac{1 \cdot 3 \cdot 5 \cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n}$$
To solve this problem, I was thinking that I would let $\cos(t)= \frac{e^{it} + e^{-it}}{2}$, then the integral will have the form: 
$$\frac{1}{2\pi}\int_0^{2\pi} \left (\frac{e^{it} + e^{-it}}{2} \right)^{2n}dt$$
From this point, I was stuck. So, would anyone please help me to walk through this problem. 
 A: You were heading in the correct direction.  Using the binomial theorem and the orthogonality of the basis functions $e^{imx}$ on $[0,2\pi]$, we have
$$\begin{align}
\frac{1}{2\pi}\int_0^{2\pi}\cos^{2n}x\,dx&=\frac{1}{2\pi}\int_0^{2\pi}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{2n}\,dx\tag 1\\\\
&=\frac{1}{2\pi}\frac{1}{2^{2n}}\,\sum_{k=0}^{2n}\binom{2n}{k}\int_0^{2\pi}e^{i(2k-2n)x}\,dx \tag 2\\\\
&=\frac{1}{2^{2n}}\binom{2n}{n}\tag 3\\\\
&=\frac{1\cdot 2\cdot 3\cdot 4 \cdots (2n-3)(2n-2)(2n-1)(2n)}{(2\cdot 4\cdot 6\cdots (2n-2)(2n))^2}\\\\
&=\frac{1\cdot 2\cdot 3\cdots (2n-1)}{2\cdot 4\cdot 6\cdots (2n-2)(2n)}\\\\
&=\frac{(2n-1)!!}{(2n)!!}
\end{align}$$
as was to be shown!
In arriving at $(1)$ we used Euler's Identity $\cos x = \frac{e^{ix}+e^{-ix}}{2}$
In going from $(1)$ to $(2)$, we used the binomial expansion $(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k}$ with $n\to 2n$, $x\to e^{ix}$, and $y\to e^{-ix}$.
In going from $(2)$ to $(3)$, we used the fact that $\int_0^{2\pi}e^{i(2k-2n)}\,dx=0$ for $k\ne n$ and $\int_0^{2\pi}e^{i(2k-2n)}\,dx=2\pi$ for $k=n$.
A: you can see that for $k \in \mathbb{Z}$ the integral of $e^{ikt}$ on $[0,2\pi]$ is zero unless $k=0$.
so from a binomial expansion of your integrand we obtain $\binom{2n}{n}2^{-2n}$
a little manipulation with the resulting fraction will give your result.
A: This should be a direct consequence of
$$
\cos^{2n}(x)
= \frac{1}{2^{2n}} \left[
\binom{2n}{n} + \sum\limits_{k=0}^{n-1} 2 \binom{2n}{k} \cos((2n-2k)x)
\right],
$$
which can be shown via induction and the equation
$$
\cos(\alpha)\cos(\beta)
= \frac{1}{2}\left[
\cos(\alpha-\beta) + \cos(\alpha+\beta)
\right].
$$
A: In arriving at  we used Euler's identity
Because here you have cos in 2 dimensional function but that identity stands for cosh which is in three dimensional I must suggest you to open the cos(power) 2n , which is equal to (1+ cos (2n))/2n and you will arrive on your answer
A: For $\gamma:[0,2 \pi] \rightarrow \mathbb{C}, \gamma(t):=\mathrm{e}^{\mathrm{i} t}$ and $f \in C\left(\partial U_{1}(0)\right)$ there is this nice equality
$$
\int_{0}^{2 \pi} f\left(\mathrm{e}^{\mathrm{i} t}\right) d t=\int_{\gamma} \frac{f(z)}{\mathrm{i} z} d z
$$
wich makes the computation after writing $\cos^{2n}(x)$ in terms of exponential functions much easier.
$
f(z)
$
becomes
$$
f(z) = \left(z+\frac{1}{z}\right)^{2n} = \frac{(z^2+1)^{2n}}{z^{2n}}
$$
Continuing from here with the binomial theorem is a little bit easier to see (it was for me atleast) and you then get your result with the Residue theorem.
