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

## Theorem

$\displaystyle \int \frac {\d x} {p^2 + q^2 \sin^2 a x} = \frac 1 {a p \sqrt {p^2 + q^2} } \arctan \frac {\sqrt {p^2 + q^2} \tan a x} p + C$

where $C$ is an arbitrary constant.

## Proof

 $\displaystyle \int \frac {\d x} {p^2 + q^2 \sin^2 a x}$ $=$ $\displaystyle \int \frac {\sec^2 a x \rd x} {p^2 \sec^2 a x + q^2 \tan^2 a x}$ multiplying numerator and denominator by $\sec^2 a x$ $\displaystyle$ $=$ $\displaystyle \int \frac {\sec^2 a x \rd x} {p^2 + \paren {p^2 + q^2} \tan^2 a x}$ Difference of Squares of Secant and Tangent $\displaystyle$ $=$ $\displaystyle \int \frac {\paren {\tan a x}' \rd x} {a p^2 + a \paren {p^2 + q^2} \tan^2 a x}$ Derivative of Tangent Function $\displaystyle$ $=$ $\displaystyle \int \frac {\d t} {a p^2 + a \paren {p^2 + q^2} t^2}$ substituting $t = \tan a x$ $\displaystyle$ $=$ $\displaystyle \frac 1 {a \paren {p^2 + q^2} } \int \frac {\rd t} {\paren {\tfrac p {\sqrt {p^2 + q^2} } }^2 + t^2}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {a \paren {p^2 + q^2} } \frac {\sqrt {p^2 + q^2} } p \, \map \arctan {\frac {\sqrt {p^2 + q^2} } p t} + C$ Primitive of $\dfrac 1 {x^2 + a^2}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} } p t} + C$ $\displaystyle$ $=$ $\displaystyle \frac 1 {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} \tan a x} p} + C$ substituting $t = \tan a x$

$\blacksquare$