<s>[INST] Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results that the integral is $0$ (as it should, since it is not $1-1$ function in this interval) I got down down the way of contour integration. Hence:
$$\begin{aligned}
\int_{0}^{2\pi}\frac{dx}{a+ b\cos x} &\overset{x=i \ln u}{=\! =\! =\!} \oint \limits_{|z|=1} \frac{dz}{iz \left [ a + \frac{b}{2}\left ( z+z^{-1} \right ) \right ]} \\ 
 &= \frac{1}{i} \oint \limits_{|z|=1} \frac{dz}{za + \frac{bz^2}{2}+\frac{b}{2}}\\ 
 &=\frac{2}{i} \oint \limits_{|z|=1} \frac{dz}{bz^2 +b +2za} \\ 
 &= \frac{2}{i} 2\pi i \sum_\text{residues} f(z)\\ 
 &= \frac{4\pi}{1+ \sqrt{1-8ab}+2a}
\end{aligned}$$
However, judging by intuition this must not be the result. Because this one restricts the integral too much. What i mean is, that for $a=6, b=3$ we have that:
$$\int_0^{2\pi} \frac{dx}{6+3\cos x}= \frac{2\pi}{3\sqrt{3}}$$
My formula cannot derive the result because then radical would be negative. What am I doing wrong here? 
 [/INST] $$\frac 1i \frac{2}{2 a z+b z^2+b}$$
This has a residue at
$$\frac{\sqrt{a^2-b^2}-a}{b}$$
Which is evaluated as
$$\frac{1}{i\sqrt{a^2-b^2}}$$
And the final answer is
$$\frac{2 \pi}{\sqrt{a^2-b^2}}$$
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