Primitive of Reciprocal of x by Root of x squared minus a squared/Arcsine Form

Theorem

$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = -\frac 1 a \arcsin \size {\frac a x} + C$

for $0 < a < \size x$.

We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:

$x > a$

or:

$x < -a$

where it is assumed that $a > 0$.

Hence:

$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} }$

$=$

$\ds \frac 1 a \arccos \size {\frac a x} + C$

$\ds $

$=$

$\ds \dfrac \pi 2 - \frac 1 a \arcsin \size {\frac a x} + C$

$\ds $

$=$

$\ds -\frac 1 a \arcsin \size {\frac a x} + C$

subsuming $\dfrac \pi 2$ into arbitrary constant $C$

$\blacksquare$