# Primitive of Reciprocal of q plus p by Secant of a x

## Theorem

$\displaystyle \int \frac {\mathrm d x} {q + p \sec a x} = \frac x q - \frac p q \int \frac {\mathrm d x} {p + q \cos a x} + C$

## Proof

 $\displaystyle \int \frac {\mathrm d x} {q + p \sec a x}$ $=$ $\displaystyle \frac 1 q \int \frac {q \ \mathrm d x} {q + p \sec a x}$ multiplying top and bottom by $q$ $\displaystyle$ $=$ $\displaystyle \frac 1 q \int \frac {\left({q + p \sec a x - p \sec a x}\right) \ \mathrm d x} {q + p \sec a x}$ $\displaystyle$ $=$ $\displaystyle \frac 1 q \int \frac {\left({q + p \sec a x}\right) \ \mathrm d x} {q + p \sec a x} - \frac p q \int \frac {\sec a x \ \mathrm d x} {q + p \sec a x}$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \frac 1 q \int \mathrm d x - \frac p q \int \frac {\sec a x \ \mathrm d x} {q + p \sec a x}$ simplifying $\displaystyle$ $=$ $\displaystyle \frac x q - \frac p q \int \frac {\sec a x \ \mathrm d x} {q + p \sec a x} + C$ Primitive of Constant $\displaystyle$ $=$ $\displaystyle \frac x q - \frac p q \int \frac {\cos a x \, \sec a x \ \mathrm d x} {q \cos a x + p \cos a x \, \sec a x} + C$ multiplying top and bottom by $\cos a x$ $\displaystyle$ $=$ $\displaystyle \frac x q - \frac p q \int \frac {\mathrm d x} {p + q \cos a x} + C$ Secant is Reciprocal of Cosine

$\blacksquare$