# Definite Integral of Reciprocal of Root of a Squared minus x Squared

## Theorem

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

## Proof

 $\displaystyle \int_0^x \frac {\d t} {\sqrt{1 - t^2} }$ $=$ $\displaystyle \sqbrk {\arcsin \frac t 1}_0^x$ Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$, Definition of Definite Integral $\displaystyle$ $=$ $\displaystyle \arcsin x - \arcsin 0$ $\displaystyle$ $=$ $\displaystyle \arcsin x$

$\blacksquare$