Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x

Theorem

 * $\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$

where $a$ and $b$ are real numbers with $a > b > 0$.

Also see

 * Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \sin x}$