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.
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 $k=n$.
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.
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 $$ \cos(\alpha)\cos(\beta) = \frac{1}{2}\left[ \cos(\alpha-\beta) + \cos(\alpha+\beta) \right]. $$
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
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.
Wallis integralsin a search engine. – Did Jul 19 '15 at 08:05