How to integrate $\int_0^{2\pi} \cos^{10}\theta \mathrm{d}\theta$ using complex analysis. I am asked to evaluate the following integral:
$$\int_0^{2\pi} \cos^{10}\theta \mathrm{d}\theta$$
I am using complex analysis. Setting $z = e^{i\theta}$, I get from Eulers formula:
$$\cos \theta = \frac{1}{2}\left(e^{i\theta} + e^{-i\theta}\right) = \frac{1}{2}\left(z + z^{-1}\right)$$
Now as $\theta$ goes from $0$ to $2\pi$, $z = e^{i\theta}$ goes one time around the unit circle. Therefore the problem is reduced to the following contour integral:
$$\oint_{C} \left(\frac{1}{2}(z + z^{-1})\right)^{10} \frac{dz}{iz}$$ where C is the unit circle.
At this point, I don't know how to move forward. I am pretty sure I am to apply the residue theorem, and the function I am integrating clearly has a pole at $z = 0$. But I don't know how to calculate that residue, since the pole is of the 10th order. Is there another approach I should take, maybe find the Laurent series of the function?
Any help is greatly appreciated!
 A: Finding the residue of the meromorphic  function
$$
f(z):=\frac{(z+z^{-1})^{10}}{2^{10}iz}
=\frac{1}{2^{10}i}\frac{(z^2+1)^{10}}{z^{11}}.\tag{1}
$$
is not difficult. 

But i don't know how to calculate the residue of that, since the pole is of the 10th order.

You are probably thinking about this formula for calculating residue at poles. But it is unnecessary here. 
All you need is to find out the coeeficient of $z^{-1}$ in (1), which means you want the coefficient of $z^{10}$ in $(z^2+1)^{10}$. By the binomial theorem, one has
$$
\frac{10!}{5!5!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2}=9\cdot 4\cdot 7.
$$
Hence the residue at $0$ is
$$
\frac{63}{2^{8}i}
$$
and by the residue theorem, the value of the integral is thus
$$
2\pi i\cdot \frac{63}{2^{8}i}=\frac{63\pi}{128}.
$$

[Added:] Without complex analysis, one can still calculate the integral in just a few steps using the recursive formula of calculating $\int \cos^nx\,dx$ ($n>0$) and taking the advantage that we are integrating over the interval $[0,2\pi]$:
$$
\begin{align}
\int_{0}^{2\pi}\cos^{10}x\,dx 
&= \frac{9}{10}\int_{0}^{2\pi}\cos^{8}x\,dx \\
&= \frac{9}{10}\frac{7}{8}\int_{0}^{2\pi}\cos^{6}x\,dx\\
&= \frac{9}{10}\frac{7}{8}\frac{5}{6}\int_{0}^{2\pi}\cos^{4}x\,dx\\
&= \frac{9}{10}\frac{7}{8}\frac{5}{6}
   \frac{3}{4}\int_{0}^{2\pi}\cos^{2}x\,dx\\
&= \frac{9}{10}\frac{7}{8}\frac{5}{6}
   \frac{3}{4} \pi=\frac{63\pi}{128}.
\end{align}
$$
A: Hint
Set $$f(z)=\frac{(z+z^{-1})^{10}}{2^{10}iz}=\frac{(z^2+1)^{10}}{2^{10}iz^{11}}.$$
Then, $$f(z)=\frac{1}{iz^{11}}\sum_{k=0}^{10}\binom{10}{k}z^{2k}=...+\frac{1}{i2^{10}z}\binom{10}{5}+...$$
