Primitive of Reciprocal of square of p plus q by Cosine of a x

Theorem

 * $\ds \int \frac {\d x} {\paren {p + q \cos a x}^2} = \frac {q \sin a x} {a \paren {q^2 - p^2} \paren {p + q \cos a x} } - \frac p {q^2 - p^2} \int \frac {\d x} {p + q \cos a x}$

Proof
Hence the result.

Also see

 * Primitive of $\dfrac 1 {\paren {p + q \sin a x}^2}$