Primitive of Reciprocal of x by Root of a squared minus x squared/Inverse Hyperbolic Secant Form

Theorem
For $a > 0$ and $0 < \size x < a$:
 * $\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {\size x} a} + C$

Proof
We note that $\sech^{-1} \dfrac x a$ is defined for $x \in \hointl 0 a$.

Hence we treat the two cases $x > 0$ and $x < 0$ separately.

First let $x > 0$.

Note that $\dfrac 1 {x \sqrt {a^2 - x^2} }$ is not defined at $\pm a$, so we are concerned only about the interval $\openint 0 a$.

Then:

Now let $x < 0$.

Let $z = -x$.

Then:

So when $x > 0$:
 * $\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac x a} + C$

and when $x < 0$:
 * $\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {-x} a} + C$

It follows that for $\size x \in \openint 0 a$:
 * $\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {\size x} a} + C$

by definition of absolute value.

Also see

 * Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$
 * Primitive of Reciprocal of $x \sqrt {x^2 - a^2}$