Primitive of Reciprocal of p plus q by Cotangent of a x

Theorem

 * $\displaystyle \int \frac {\d x} {p + q \cot a x} = \frac {p x} {p^2 + q^2} - \frac q {a \paren {p^2 + q^2} } \ln \size {p \sin a x + q \cos a x} + C$

Proof
We have:


 * $\dfrac \d {\d x} \paren {p \sin a x + q \cos a x} = a p \cos a x - a q \sin a x$

Thus:

Also see

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


 * Primitive of $\dfrac 1 {p + q \cos a x}$


 * Primitive of $\dfrac 1 {p + q \tan a x}$


 * Primitive of $\dfrac 1 {q + p \sec a x}$


 * Primitive of $\dfrac 1 {q + p \csc a x}$