Primitive of Reciprocal of p squared minus square of q by Cosine of a x

Theorem

 * $\ds \int \frac {\rd x} {p^2 - q^2 \cos^2 a x} = \begin {cases}

\dfrac 1 {a p \sqrt {p^2 - q^2} } \arctan \dfrac {p \tan a x} {\sqrt {p^2 - q^2} } & : p^2 > q^2 \\ \dfrac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\dfrac {p \tan a x - \sqrt {q^2 - p^2} } {p \tan a x + \sqrt {q^2 - p^2} } } & : p^2 < q^2 \end {cases}$

Proof
Let:

There are two cases to address.

First, suppose $p^2 > q^2$.

Then we have that $p^2 - q^2 > 0$, and so:

Now suppose $p^2 < q^2$.

Then we have that $p^2 - q^2 < 0$, and so:

Also see

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