Primitive of x squared by Arcsecant of x over a

Theorem

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

\dfrac {x^3} 3 \arcsec \dfrac x a - \dfrac {a x \sqrt {x^2 - a^2} } 6 - \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\ \dfrac {x^3} 3 \arcsec \dfrac x a + \dfrac {a x \sqrt {x^2 - a^2} } 6 + \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^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:

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 $x^2 \arcsin \dfrac x a$


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


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


 * Primitive of $x^2 \arccot \dfrac x a$


 * Primitive of $x^2 \arccsc \dfrac x a$