# 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:

 $\displaystyle u$ $=$ $\displaystyle \operatorname{arcsec} \frac x a$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle \begin{cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \operatorname{arcsec} \dfrac x a < \dfrac \pi 2 \\ \dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \operatorname{arcsec} \dfrac x a < \pi \\ \end{cases}$ Derivative of $\operatorname{arcsec} \dfrac x a$

and let:

 $\displaystyle \frac {\mathrm d v} {\mathrm d x}$ $=$ $\displaystyle \frac 1 {x^2}$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {-1} x$ Primitive of Power

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

Then:

 $\displaystyle \int \frac {\operatorname{arcsec} \frac x a} {x^2}$ $=$ $\displaystyle \frac {-1} x \operatorname{arcsec} \frac x a - \int \frac {-1} x \left({\frac a {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {-\operatorname{arcsec} \frac x a} x + a \int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } + C$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac {-\operatorname{arcsec} \frac x a} x + a \left({\frac {\sqrt {x^2 - a^2} } {a^2 x} }\right) + C$ Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$ $\displaystyle$ $=$ $\displaystyle \frac {-\operatorname{arcsec} \frac x a} x + \frac {\sqrt{x^2 - a^2} } {a x} + C$ simplifying

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

Then:

 $\displaystyle \int \frac {\operatorname{arcsec} \frac x a} {x^2}$ $=$ $\displaystyle \frac {-1} x \operatorname{arcsec} \frac x a - \int \frac {-1} x \left({\frac {-a} {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {-\operatorname{arcsec} \frac x a} x - a \int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } + C$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac {-\operatorname{arcsec} \frac x a} x - a \left({\frac {\sqrt {x^2 - a^2} } {a^2 x} }\right) + C$ Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$ $\displaystyle$ $=$ $\displaystyle \frac {-\operatorname{arcsec} \frac x a} x - \frac {\sqrt{x^2 - a^2} } {a x} + C$ simplifying

$\blacksquare$