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

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

$\blacksquare$