How to integrate a complex function which doesn't have any singularities? I am trying to solve this question. 
$$\int_{0}^{2\pi}(\cos(z))^{6}dz$$
Usually when I try doing complex integration questions the function has a singularity and you are given a simple closed curve. So I am able to use Cauchy's Residue theorem or Cauchy's integral formula to evaluate the integral. The problem I'm having which this one is that they haven't given a function that has singularities and it isn't a simple closed curve either. I haven't run into this type of problem before and I'm not even sure where to begin. I was wondering whether someone can help on how to go about doing these types of questions. Sorry about not being able to show work I'm not sure where to start. 
 A: Firstly we can put: $\cos(z)=(e^{iz}+e^{-iz})/2$ ,
$$
\int_{0}^{2\pi}(\frac{e^{iz}+e^{-iz}}{2})^{6}dz
$$
$$
\frac{1}{2^6}\int_{0}^{2\pi}(e^{iz}+e^{-iz})^{6}dz
$$
$$
\frac{1}{2^6}\int_{0}^{2\pi}((e^{iz})^6+6(e^{iz})^5e^{-iz}+15(e^{iz})^4(e^{-iz})^2+20(e^{iz})^3(e^{-iz})^3+15(e^{iz})^2(e^{-iz})^4+6(e^{iz})(e^{-iz})^5+(e^{-iz})^6)dz
$$
$$
\frac{1}{2^6}\int_{0}^{2\pi}((e^{iz})^6+6(e^{iz})^4+15(e^{iz})^2+20+15(e^{-iz})^2+6(e^{-iz})^4+(e^{-iz})^6)dz
$$
"Now for first term $\frac{1}{2^6}\int_{0}^{2\pi}(e^{iz})^6dz$ , put $t=e^{iz}$ so $dt=i.e^{iz}dz$. 
Hence, first term evaluates to (with changed limits)  $\frac{1}{i.2^6}\int_{0}^{1}t^5dt$ , which is equal to $\frac{1}{i.6.2^6}$ ."
Finally using this similar technique,
$$
\frac{1}{i.6.2^6}+\frac{6}{i.4.2^6}+\frac{15}{i.2.2^6}+\frac{40\pi}{2^6}+\frac{-15}{i.2.2^6}+\frac{-6}{i.4.2^6}+\frac{-1}{i.6.2^6}
$$
$$
\frac{5\pi}{8}
$$
Hence, the final answer is $\frac{5\pi}{8}$ .
A: Without any specification of an integration path, there is no reason to interpret the integral as anything more than it's face value. In other words, just ignore the fact that $z$ is complex. Just as @SahilKumar has shown elsewhere on this page, the integral is simply $5\pi/8$, just as if you integrated $\cos^6(\theta)$.
This is a good time to remind ourselves of the well-known solution for
$$
\int_0^{2\pi}\cos^n\theta\ d\theta=\frac{\pi}{2^{n-1}}
\left(
\begin{matrix}
n\\
n/2
\end{matrix}
\right)=\frac{\pi}{2^{n-1}}\frac{n!}{(n/2)!^2}
\ \ \ \text{for } n \text{ even}, \text{ and 0 (zero) otherwise}
$$
