Primitive of Arcsecant of x over a over x squared

Theorem

 * $\ds \int \dfrac 1 {x^2} \arcsec \frac x a \rd x = \begin {cases}

-\dfrac 1 x \arcsec \dfrac x a + \dfrac {\sqrt {x^2 - a^2} } {a x} + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ -\dfrac 1 x \arcsec \dfrac x a - \dfrac {\sqrt {x^2 - a^2} } {a x} + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end {cases}$

Proof
With a view to expressing the primitive in the form:
 * $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

and let:

First let $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.

Then:

Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.

Then:

Also see

 * Primitive of $\dfrac 1 {x^2} \arcsin \dfrac x a$


 * Primitive of $\dfrac 1 {x^2} \arccos \dfrac x a$


 * Primitive of $\dfrac 1 {x^2} \arctan \dfrac x a$


 * Primitive of $\dfrac 1 {x^2} \arccot \dfrac x a$


 * Primitive of $\dfrac 1 {x^2} \arccsc \dfrac x a$