Definite Integral from 0 to a of Reciprocal of Root of a Squared minus x Squared/Proof 2

Theorem

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

for $a > 0$.

Proof

\(\ds \int_0^a \frac {\d x} {\sqrt {a^2 - x^2} }\)

\(=\)

\(\ds \frac {a^{1 - \frac 2 2} \map \Gamma {\frac 1 2} \map \Gamma {-\frac 1 2 + 1} } {2 \map \Gamma {\frac 1 2 - \frac 1 2 + 1} }\)

Definite Integral from 0 to a of $x^m \paren {a^n - x^n}^p$

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \times 0!} \paren {\map \Gamma {\frac 1 2} }^2\)

Gamma Function Extends Factorial

\(\ds \)

\(=\)

\(\ds \frac 1 2 \paren {\sqrt \pi}^2\)

Gamma Function of One Half

\(\ds \)

\(=\)

\(\ds \frac \pi 2\)

$\blacksquare$