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

Theorem

 * $\displaystyle \int_0^{2 \pi} \frac {\d x} {\paren {a + b \cos x}^2} = \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }$

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

Proof
From Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \cos x}$, we have:


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

We have:

and:

giving:


 * $\displaystyle \int_0^{2 \pi} \frac {\d x} {\paren {a + b \cos x}^2} = \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }$

Also see

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