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

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Theorem

$\displaystyle \int \frac {\mathrm d x} {p^2 - q^2 \cos^2 a x} = \begin{cases} \displaystyle \frac 1 {a p \sqrt{p^2 - q^2} } \arctan \frac {p \tan a x} {\sqrt{p^2 - q^2} } & : p^2 > q^2 \\ \displaystyle \frac 1 {2 a p \sqrt{q^2 - p^2} } \ln \left\vert{\frac {p \tan a x - \sqrt{q^2 - p^2} } {p \tan a x + \sqrt{q^2 - p^2} } }\right\vert & : p^2 < q^2 \end{cases}$


Proof


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