Primitive of Arcsecant of x over a over x squared

Theorem

 * $\displaystyle \int \frac {\operatorname{arcsec} \frac x a} {x^2} \ \mathrm d x = \begin{cases}

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

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

let:

and let:

First let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({0 \,.\,.\,\dfrac \pi 2}\right)$.

Then:

Similarly, let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.

Then:

Also see

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


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


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


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


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