Evaluating $\int_0^{2\pi}\frac{4 \cdot \cos(x)}{5-4 \cdot \cos(x)}dx=\frac{4}{3}\pi\;$ with residue? Wolfam Alpha says that $$\int_0^{2\pi}\frac{4 \cdot \cos(x)}{5-4 \cdot \cos(x)}dx=\frac{4}{3}\pi$$
I want to calculate this on my own, but have problems.Would be great, if you could tell me where and what it is.
$$\begin{align}\int_0^{2\pi}\frac{4 \cdot \cos(x)}{5-4 \cdot \cos(x)}dx & =\operatorname{Re} \Biggr(\int_0^{2\pi}\frac{4e^{ix}}{5-4e^{ix}}dx \Biggl) \\ 
& = \operatorname{Re} \Biggr(\int_0^{2\pi}\frac{4z}{(5-4z)iz}dz \Biggl)\\
& = \operatorname{Re} \Biggr(-i\int_0^{2\pi}\frac{4}{(5-4z)}dz \Biggl) \\
& = \operatorname{Re} \Biggr(-i\Bigg[\ln(5-4z)\Bigg]_0^{2\pi}\Bigg) \\
& = \operatorname{Re} \Biggr(-i\ln(\frac{5-8\pi}{5})\Bigg)\\
\end{align}$$
I guess that I can't calculate the integral in this way, right ? 
I guess that I can't calculate the integral in this way (4), right ? 

This is not the right way of calculating.$$\int_0^{2\pi}\frac{4 \cdot \cos(x)}{5-4 \cdot \cos(x)}dx=\int_{|z|=1}\frac{2z+2z^{-1}}{5-2z-2z^{-1}}\dfrac{dz}{iz}=\int_{|z|=1}\frac{2z^2+2}{5z-2z^2-2}\dfrac{dz}{iz}$$The singularities are in $2,\dfrac{1}{2},0$ where $$\operatorname{Re}z_0=-1\\\operatorname{Re}z_{\dfrac{1}{2}}=\dfrac{5}{3}$$therefore $$I=\dfrac{1}{i}2\pi i\left(-1+\dfrac{5}{3}\right)=\pi\dfrac{4}{3}$$
