Proving $\int_{0}^{2 \pi}\cos^{2n}(t)dt=2\pi\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$ I have just evaluated $$\int_{|z|=1}\frac{(z^2+1)^{2n}}{z^{2n+1}}dz$$ to be $2 \pi i$. I now want to use this fact to show that $$\int_{0}^{2 \pi}\cos^{2n}(t)dt=2\pi\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdot8\cdots(2n)}.$$
I thought perhaps the next course of action would be to show the Taylor series representation of $\cos^{2n}(x)$ is $$\sum(-1)^k \frac{2^{2k-1}}{2k!}\cdot x^{2k}$$ then rearrange this somehow to show that you can give the first integral I mentioned in terms of this Taylor expansion and then compare it to the binomial expanded version of the first integral.
Will this work or am I doing it wrong ? any hints or tips are greatly appreciated in advance.
P.S. 
This is an assignment question so please no full answers.
 A: This probably will not work because you can't quite evaluate the integral over an infinite sum (there are convergence issues you have to deal with) and that the Taylor series you have doesn't seem to be too similar to the expression in the first integral.
You can certainly try, but I would personally work from the first integral to try to get the second.
In the first place, I think you have actually incorrectly evaluated the first integral.
$$\frac{(z^2+1)^{2n}}{z^{2n+1}}=\frac{z^{4n}+2nz^{4n-2}+\cdots + C_{n}^{2n}z^{2n}+\cdots + 1}{z^{2n}}=z^{2n-1}+\cdots +C_{n}^{2n}z^{-1} + \cdots + z^{-2n}$$
The function $f(z)=\dfrac{(z^2+1)^{2n}}{z^{2n+1}}$ only has singularity at $z=0$ at which $\text{res}(f;0)=C_{n}^{2n}$ as shown in the calculation above (simply the coefficient of the $z^{-1}$ term).
Hence $$\int_{|z|=1}\frac{(z^2+1)^{2n}}{z^{2n+1}}dz=2\pi i \cdot \text{res}(f;0) = 2\pi i C_{n}^{2n}=2\pi i \frac{(2n)\cdot (2n-1) \cdot \cdots \cdot (n+1)}{(n) \cdot (n-1) \cdot \cdots \cdot (1)}$$
On the other hand, you can also evaluate this integral by parameterizing it.
Hint:
Suppose we parameterize a path $\gamma$ with $\gamma (t)$ where $t \in [a,b]$. Then
$$\int_{\gamma}f(z)dz=\int_{t=a}^{t=b}f\bigl(\gamma (t) \bigr)\gamma ' (t)dt$$
A: You are looking for 
$$\int_0^{2\pi} \cos^{2m}\theta\,d\theta=4\int_0^{\pi/2} \cos^{2m}\theta\,d\theta \qquad \text{for } m\in\mathbb{N}$$
Now, perfmorm the substitution $x=\tan \theta$.
Our integral is then
$$4\int_0^{\infty}\cos^{2m}\arctan x\cdot \frac{dx}{1+x^2}$$
Using the trigonometric identity, 
$$1+\tan^2 \theta=\sec^2\theta$$
We can solve for $\cos^2 \theta$:
$$\cos^{2}\theta=\frac{1}{1+\tan^2 \theta}$$
And substituting $\arctan x=\theta$, we find that
\begin{align}
\cos^{2m}\arctan x=\left(\frac{1}{1+x^2}\right)^m
\end{align}
Our original integral now reduces to 
$$4\int_0^{\infty} dx\frac{1}{\left(1+x^2\right)^{m+1}}$$
There are many different ways to evaluate this integral, but I will use the Beta function. Substituting $u=x^2$, our integral is now
$$2\int_0^{\infty} \frac{u^{-1/2}}{\left(1+u\right)^{m+1}}\,du$$
Using the definition of the Beta function, this integral is equal to
\begin{align}
2\,\mathcal{B}\left(1/2,m+1/2\right)&=2\frac{\Gamma(1/2)\Gamma(m+1/2)}{\Gamma(m+1)}\\
&=2\pi\frac{\Gamma(2m)}{\Gamma(m)\Gamma(m+1)} 2^{1-2m} \\
&= 2\pi 2^{1-2m} \frac{(2m-1)!}{(m!)(m-1)!} \\
&= \frac{\pi}{4^{m-1}} {{2m}\choose {m}}
\end{align}
