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

Theorem

 * $\displaystyle \int \frac {1}{\sqrt{a^2 - x^2}} \ \mathrm dx = \arcsin{\frac{|x|}{a}} + C$

where $a$ is a strictly positive constant and $a^2 > x^2$.

Proof
Substitute:


 * $\sin \theta = \dfrac {x}{a} \iff x = a \sin \theta$

for $\theta \in \left(-\dfrac {\pi}{2}..\dfrac {\pi}{2}\right)$.

From Boundedness of Sine and Cosine, this substitution is valid for all $x/a \in \left(-1..1\right)$.

By hypothesis:

$\iff -1 < \left(x/a \right) < 1$

... so this substitution will not change the domain of the integrand.

By the definition of absolute value:


 * $ \ \displaystyle \int \frac {\cos \theta} {\left \vert \cos \theta \right \vert} \ \mathrm d \theta = \begin{cases}

\int \mathrm d \theta & : \cos \theta > 0 \\ \int -\mathrm d \theta & : \cos \theta < 0 \end{cases}$


 * $ = \begin{cases}

\theta + C & : \theta > 0 \\ -\theta + C & : \theta < 0 \end{cases}$


 * $ = |\theta| + C$

As $\theta$ was stipulated to be in the open interval $\left(-\dfrac {\pi}{2}..\dfrac {\pi}{2}\right)$:


 * $\sin \theta = \dfrac {x}{a} \iff \theta = \arcsin \dfrac {x}{a}$

The answer in terms of $x$, then, is: