Primitive of Function of Arcsecant

Theorem

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

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

Proof
First note that:

Then:

Also see

 * Primitive of Function of Arcsine
 * Primitive of Function of Arccosine
 * Primitive of Function of Arctangent
 * Primitive of Function of Arccotangent
 * Primitive of Function of Arccosecant