# Primitive of Power of x by Arcsecant of x over a

## Theorem

$\displaystyle \int x^m \operatorname{arcsec} \frac x a \ \mathrm d x = \begin{cases} \displaystyle \frac {x^{m + 1} } {m + 1} \operatorname{arcsec} \frac x a - \frac a {m + 1} \int \frac {x^m \ \mathrm d x} {\sqrt {x^2 - a^2} } + C & : 0 < \operatorname{arcsec} \dfrac x a < \dfrac \pi 2 \\ \displaystyle \frac {x^{m + 1} } {m + 1} \operatorname{arcsec} \frac x a + \frac a {m + 1} \int \frac {x^m \ \mathrm d x} {\sqrt {x^2 - a^2} } + C & : \dfrac \pi 2 < \operatorname{arcsec} \dfrac x a < \pi \\ \end{cases}$

## 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:

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

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

Then:

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

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

Then:

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

$\blacksquare$