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

Theorem

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

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 $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.

From Real Sine Function is Bounded and Shape of Sine Function, this substitution is valid for all $x / a \in \openint {-1} 1$.

By hypothesis:

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

Then:

We have defined $\theta$ to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.

From Sine and Cosine are Periodic on Reals, $\cos \theta > 0$ for the entire interval. Therefore the absolute value is unnecessary, and the integral simplifies to:

As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:


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

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


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

Also see

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


 * Derivative of Arcsine Function