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!
Best Answer
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.
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 coefficient 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 $[0,2\pi]$: $$ \begin{align} \int_{0}^{2\pi}\cos^{10}x\,dx &= \frac{9}{10}\int_{0}^{2\pi}\cos^{8}x\,dx \\ &= \frac{9}{10}\frac{7}{8}\int_{0}^{2\pi}\cos^{6}x\,dx\\ &= \frac{9}{10}\frac{7}{8}\frac{5}{6}\int_{0}^{2\pi}\cos^{4}x\,dx\\ &= \frac{9}{10}\frac{7}{8}\frac{5}{6} \frac{3}{4}\int_{0}^{2\pi}\cos^{2}x\,dx\\ &= \frac{9}{10}\frac{7}{8}\frac{5}{6} \frac{3}{4} \pi=\frac{63\pi}{128}. \end{align} $$