# Primitive of x by Arcsecant of x over a

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## Theorem

$\ds \int x \arcsec \frac x a \rd x = \begin{cases} \dfrac {x^2} 2 \arcsec \dfrac x a - \dfrac {a \sqrt {x^2 - a^2} } 2 + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ \dfrac {x^2} 2 \arcsec \dfrac x a + \dfrac {a \sqrt {x^2 - a^2} } 2 + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\ \end{cases}$

## 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:

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

and let:

 $\ds \frac {\d v} {\d x}$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds \frac {x^2} 2$ Primitive of Power

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

Then:

 $\ds \int x \arcsec \frac x a \rd x$ $=$ $\ds \frac {x^2} 2 \arcsec \frac x a - \int \frac {x^2} 2 \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x + C$ Integration by Parts $\ds$ $=$ $\ds \frac {x^2} 2 \arcsec \frac x a - \frac a 2 \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + C$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds \frac {x^2} 2 \arcsec \frac x a - \frac a 2 \sqrt {x^2 - a^2} + C$ Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$ $\ds$ $=$ $\ds \frac {x^2} 2 \arcsec \frac x a - \frac {a \sqrt {x^2 - a^2} } 2 + C$ simplifying

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

Then:

 $\ds \int x \arcsec \frac x a \rd x$ $=$ $\ds \frac {x^2} 2 \arcsec \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C$ Integration by Parts $\ds$ $=$ $\ds \frac {x^2} 2 \arcsec \frac x a + \frac a 2 \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + C$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds \frac {x^2} 2 \arcsec \frac x a + \frac a 2 \sqrt {x^2 - a^2} + C$ Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$ $\ds$ $=$ $\ds \frac {x^2} 2 \arcsec \frac x a + \frac {a \sqrt {x^2 - a^2} } 2 + C$ simplifying

$\blacksquare$