Definite Integral from 0 to a of Reciprocal of Root of a Squared minus x Squared/Proof 1

Theorem

$\ds \int_0^a \dfrac {\d x} {\sqrt {a^2 - x^2} } = \frac \pi 2$

for $a > 0$.

$\ds \int_0^a \dfrac {\d x} {\sqrt {a^2 - x^2} }$

$=$

$\ds \int_0^{\mathop \to a} \dfrac {\d x} {\sqrt {a^2 - x^2} }$

as $\dfrac 1 {\sqrt {a^2 - x^2} }$ does not exist for $x = a$

$\ds $

$=$

$\ds \lim_{\gamma \mathop \to a} \int_0^\gamma \dfrac {\d x} {\sqrt {a^2 - x^2} }$

$\ds $

$=$

$\ds \lim_{\gamma \mathop \to a} \intlimits {\arcsin \frac x a} 0 \gamma$

$\ds $

$=$

$\ds \lim_{\gamma \mathop \to a} \paren {\arcsin \frac \gamma a - \arcsin 0}$

$\ds $

$=$

$\ds \lim_{\gamma \mathop \to a} \arcsin \frac \gamma a$

$\ds $

$=$

$\ds \arcsin \frac a a$

$\ds $

$=$

$\ds \arcsin 1$

$\ds $

$=$

$\ds \frac \pi 2$

$\blacksquare$