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

Theorem

 * $\ds \int_0^{\pi/2} \frac 1 {a + b \cos x} \rd x = \frac 1 {\sqrt {a^2 - b^2} } \map \arccos {\frac b a}$

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

Proof
Since $a > b > 0$, we have $a^2 > b^2$.

So: