How to evaluate $\int_0^{2\pi} \frac{d\theta}{A+B\cos\theta}$? I'm having a trouble with this integral expression:
$$\int_0^{2\pi} \frac{d\theta}{A+B \cos\theta}$$
I've done this substitution: $t= \tan(\theta/2)$
and get: $\displaystyle \cos\theta= \frac{1-t^2}{1+t^2}$ and $\displaystyle d\theta=\frac{2}{1+t^2}dt$ where $\displaystyle \cos^2\theta/2=\frac{1}{1+t^2}$
then the integral becomes:
$$\int\frac{2 \, dt}{(A-B)t^2+(A+B)}= \sqrt\frac{A+B}{A-B} \arctan \left(\left(\sqrt\frac{A+B}{A-B} \right) t\right)$$ 
However, I'm not sure about the new limits since $\tan$ has period $\pi$ so what I have to do at this point to decide the new limits? And of course if you find some mistake in what I've done before, please let me know!

To avoid confusion with limits, note that
$$\int_0^{\pi} \dfrac{dx}{a+b \cos(x)} = \int_{\pi}^{2\pi} \dfrac{dx}{a+b \cos(x)}$$
Hence, we have
$$I = \int_0^{2\pi} \dfrac{dx}{a+b \cos(x)} = 2\int_0^{\pi} \dfrac{dx}{a+b \cos(x)}$$
Now use your substitution $t = \tan(x/2)$ and note that $t$ goes from $0$ to $\infty$ as $x$ goes from $0$ to $\pi$.
EDIT
An easier way if you are familiar with a little bit of complex analysis is as follows.
Let $z = e^{ix}$. Then we have that $dz = iz dx$. Hence,
\begin{align}
I & = \int_0^{2 \pi} \dfrac{dx}{a+b \cos(x)} = \oint_{\vert z \vert = 1} \dfrac{dz}{iz \left( a + \dfrac{b}2 \left(z+ \dfrac1z \right)\right)}\\
& = \dfrac1i \oint_{\vert z \vert = 1} \dfrac{dz}{az + \dfrac{b}2 \left(z^2+1 \right)} = \dfrac2{ib} \oint_{\vert z \vert = 1}\dfrac{dz}{z^2 + \dfrac{2a}b z + 1}\\
& = \dfrac2{ib} \oint_{\vert z \vert = 1} \dfrac{dz}{\left(z + \dfrac{a}b \right)^2 + 1 - \dfrac{a^2}{b^2}} = \dfrac2{ib} \oint_{\vert z \vert = 1} \dfrac{dz}{\left(z + r + \sqrt{r^2-1} \right)\left(z + r - \sqrt{r^2-1} \right)}
\end{align}
where $r = \dfrac{a}b \in \mathbb{R}$.
First note that if $\vert r \vert \leq 1$, then $\left \vert r \pm \sqrt{r^2-1} \right \vert = 1$. Hence, the integral doesn't exist.
If $r > 1$, then $ \left \vert r + \sqrt{r^2 - 1} \right \vert > 1$ and $\left \vert r - \sqrt{r^2-1} \right \vert < 1$. Hence, by Cauchy integral formula, we have that
$$I = \dfrac{2}{ib} 2 \pi i \dfrac1{2 \sqrt{r^2-1}} = \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}}$$
If $r < -1$, then $\left \vert r - \sqrt{r^2 - 1} \right \vert > 1$ and $\left \vert r + \sqrt{r^2-1} \right \vert < 1$. Hence, by Cauchy integral formula, we have that
$$I = \dfrac{2}{ib} 2 \pi i \dfrac{-1}{2 \sqrt{r^2-1}} = \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}}$$
Hence, to summarize,
$$I = \begin{cases} \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}} & \text{if } \vert a \vert > \vert b\\ \text{Does not exists} & \text{if } \vert a \vert \leq \vert b\end{cases}$$
