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

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {\left({p + q \sin a x}\right)^2} = \frac {q \cos a x} {a \left({p^2 - q^2}\right) \left({p + q \sin a x}\right)} + \frac p {p^2 - q^2} \int \frac {\mathrm d x} {p + q \sin a x}$