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

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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 $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2\) $\quad$ $\quad$

$\blacksquare$


Sources