# Primitive of Function of Arcsecant

## Theorem

$\ds \int \map F {\arcsec \frac x a} \rd x = a \int \map F u \sec u \tan u \rd u$

where $u = \arcsec \dfrac x a$.

## Proof

First note that:

 $\ds u$ $=$ $\ds \arcsec \frac x a$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds a \sec u$ Definition of Arcsecant

Then:

 $\ds u$ $=$ $\ds \arcsec \frac x a$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds \frac a {\size x {\sqrt {x^2 - a^2} } }$ Derivative of Arcsecant Function: Corollary 1 $\ds \leadsto \ \$ $\ds \int \map F {\arcsec \frac x a} \rd x$ $=$ $\ds \int \map F u \, \frac {\size x {\sqrt {x^2 - a^2} } } a \rd u$ Primitive of Composite Function $\ds$ $=$ $\ds \int \map F u \frac {\size {a \sec u} {\sqrt {a^2 \sec^2 u - a^2} } } a \rd u$ Definition of $x$ $\ds$ $=$ $\ds \int \map F u \size {\sec u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u$ $\ds$ $=$ $\ds \int \map F u a \sec u {\sqrt {\sec^2 u - 1} } \rd u$ $\sec u > 0$ in this domain $\ds$ $=$ $\ds \int \map F u a \sec u \tan u \rd u$ Difference of Squares of Secant and Tangent $\ds$ $=$ $\ds a \int \map F u \sec u \tan u \rd u$ Primitive of Constant Multiple of Function

$\blacksquare$