How to integrate a complex function which doesn't have any singularities?

Question

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.

Answer 1

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 $\frac{5\pi}{8}$ .

Answer 2

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

$$ \int_0^{2\pi}\cos^n\theta\ d\theta=\frac{\pi}{2^{n-1}} \left( \begin{matrix} n\\ n/2 \end{matrix} \right)=\frac{\pi}{2^{n-1}}\frac{n!}{(n/2)!^2} \ \ \ \text{for } n \text{ even}, \text{ and 0 (zero) otherwise} $$