Find $\int\limits_{0}^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$ I have to find $$\int_0^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$$
I tried to do it by substitution $t = \tan(\frac{x}{2})$
Then we have that $$\cos(x) = \frac{1-t^2}{1+t^2} \quad dx = \frac{2\,dt}{1+t^2}$$ but then also limits of integration are changing so we have
$$\int\limits_{t(0)}^{t(2\pi)} \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2dt}{1+t^2} = \int\limits_0^0 \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2\,dt}{1+t^2} = 0$$
I figured out that it is not correct because $\tan(\frac{\pi}{2})$ is not defined and $t(\pi) = \tan(\frac{\pi}{2})$ and $\pi \in [0, 2\pi]$. How can I "repair" that and do it right?

While there have been several clean and different approaches shown here, I'm surprised that nobody has shown that contour integration works well also.  To that end here we go....
Let $z=e^{ix}$ so that $dx=dz/iz$ and $\cos x = \frac12 (z+z^{-1})$.  We have 
$$\begin{align}
I&=\int_0^{2\pi} \frac{1}{5-3\cos x}dx\\\\
&=\oint_C \frac{1}{5-3\frac12 (z+z^{-1})}\frac{dz}{iz}\\\\
&=\oint_C \frac{\frac23 i}{(z-3)(z-1/3)}dz\\\\
&=2\pi i \text{Res}_{\text{in C}} \left(\frac{\frac23 i}{(z-3)(z-1/3)}\right)
\end{align}$$
where $C$ is the unit circle in the complex $z$-plane.  
Note that the pole at $z=1/3$ is inside the unit circle whereas the pole at $z=3$ is outside the unit circle.  
Thus, the residue is given by 
$$\text{Res}_{z=1/3} \left(\frac{\frac23 i}{(z-3)(z-1/3)}\right)=\frac23 i\,\lim_{z\to 1/3}\left((z-1/3)\,\frac{1}{(z-3)(z-1/3)}\right)=-i/4$$
which after multiplying by $2\pi i$ shows that 
$$I=\pi/2$$
as expected!
