# Definite Integral from 0 to a of Reciprocal of Root of a Squared minus x Squared

## Theorem

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

for $a > 0$.

## Proof

 $\displaystyle \int_0^a \dfrac {\d x} {\sqrt {a^2 - x^2} }$ $=$ $\displaystyle \int_0^{\mathop \to a} \dfrac {\d x} {\sqrt {a^2 - x^2} }$ $\quad$ as $\dfrac 1 {\sqrt {a^2 - x^2} }$ does not exist for $x = a$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{\gamma \mathop \to a} \int_0^\gamma \dfrac {\d x} {\sqrt {a^2 - x^2} }$ $\quad$ Definition of Improper Integral on Open Above Interval $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{\gamma \mathop \to a} \left[{\arcsin \frac x a}\right]_0^\gamma$ $\quad$ Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{\gamma \mathop \to a} \left({\arcsin \frac \gamma a - \arcsin 0}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{\gamma \mathop \to a} \arcsin \frac \gamma a$ $\quad$ Arcsine of Zero is Zero $\quad$ $\displaystyle$ $=$ $\displaystyle \arcsin \frac a a$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \arcsin 1$ $\quad$ Arcsine of One is Half Pi $\quad$ $\displaystyle$ $=$ $\displaystyle \frac \pi 2$ $\quad$ $\quad$

$\blacksquare$