# Beta Function of Half with Half/Proof 2

## Theorem

$\map \Beta {\dfrac 1 2, \dfrac 1 2} = \pi$

where $\Beta$ denotes the Beta function.

## Proof

By definition of the Beta function:

$\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$

Thus:

 $\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}$ $=$ $\ds \int_{\mathop \to 0}^{\mathop \to 1} t^{\frac 1 2 - 1} \paren {1 - t}^{\frac 1 2 - 1} \rd t$ $\ds$ $=$ $\ds \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {\rd t} {\sqrt {t \paren {1 - t} } }$

Let $t = \sin^2 \theta$.

Then:

$\rd t = 2 \sin \theta \cos \theta \rd \theta$

and:

 $\ds t$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \sin^2 \theta$ $=$ $\ds 0$ $\ds t$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds \sin^2 \theta$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds \theta$ $=$ $\ds \pi / 2$

and so:

 $\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}$ $=$ $\ds \int_{\mathop \to 0}^{\mathop \to \pi / 2} \frac {2 \sin \theta \cos \theta \rd \theta} {\sin \theta \sqrt {1 - \sin^2 \theta} }$ Integration by Substitution $\ds$ $=$ $\ds \int_{\mathop \to 0}^{\mathop \to \pi / 2} \frac {2 \sin \theta \cos \theta \rd \theta} {\sin \theta \cos \theta}$ Sum of Squares of Sine and Cosine $\ds$ $=$ $\ds \int_{\mathop \to 0}^{\mathop \to \pi / 2} 2 \rd \theta$ $\ds$ $=$ $\ds \bigintlimits {2 \theta} 0 {\pi / 2}$ $\ds$ $=$ $\ds 2 \times \pi / 2 - 0$ $\ds$ $=$ $\ds \pi$

$\blacksquare$