# Primitive of Reciprocal of Root of a squared minus x squared

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## 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 $a^2 > x^2$.

### Corollary

$\displaystyle \int_0^x \frac {\d t} {\sqrt{1 - t^2} } = \arcsin x$

## Proof

 $\displaystyle \int \frac 1 {\sqrt {a^2 - x^2} } \rd x$ $=$ $\displaystyle \int \frac {\rd x} {\sqrt {a^2 \paren {1 - \frac {x^2} {a^2} } } }$ factor $a^2$ out of the radicand $\displaystyle$ $=$ $\displaystyle \int \frac {\rd x} {\sqrt{a^2} \sqrt {1 - \paren {\frac x a}^2} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 a \int \frac {\rd x} {\sqrt {1 - \paren {\frac x a}^2} }$
$\sin \theta = \dfrac x a \iff x = a \sin \theta$

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 $x / a \in \openint {-1} 1$.

 $\displaystyle a^2$ $>$ $\displaystyle x^2$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle 1$ $>$ $\displaystyle \frac {x^2} {a^2}$ dividing both terms by $a^2$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle 1$ $>$ $\displaystyle \paren {\frac x a}^2$ Powers of Group Elements $\displaystyle \leadstoandfrom \ \$ $\displaystyle 1$ $>$ $\displaystyle \size {\frac x a}$ taking the square root of both terms $\displaystyle \leadstoandfrom \ \$ $\displaystyle -1$ $<$ $\displaystyle \paren {\frac x a} < 1$ Negative of Absolute Value

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

Then:

 $\displaystyle x$ $=$ $\displaystyle a \sin \theta$ from above $\displaystyle \leadstp \ \$ $\displaystyle 1$ $=$ $\displaystyle a \cos \theta \frac {\rd \theta} {\rd x}$ differentiating with respect to $x$, Derivative of Sine Function, Chain Rule for Derivatives $\displaystyle \frac 1 a \int \frac 1 {\sqrt {1 - \paren {\frac x a}^2 } } \rd x$ $=$ $\displaystyle \frac 1 a \int \frac {a \cos \theta} {\sqrt {1 - \sin^2 \theta} } \frac {\rd \theta} {\rd x} \rd x$ from above $\displaystyle$ $=$ $\displaystyle \frac a a \int \frac {\cos \theta} {\sqrt {1 - \sin^2 \theta} } \rd \theta$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \int \frac {\cos \theta} {\sqrt {\cos^2 \theta} } \rd \theta$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle \int \frac {\cos \theta} {\size {\cos \theta} } \rd \theta$

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:

 $\displaystyle \int \rd \theta$ $=$ $\displaystyle \theta + C$

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:

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

$\blacksquare$