Trigonometric Representation of the Beta Function
Theorem
$$ B(p,q) = 2 \int_{0}^{{\pi} \over {2}} \left( \sin \theta \right) ^{2p-1} \left( \cos \theta \right) ^{2q-1} d \theta $$
Description
No matter what kind of mathematics it is, being able to express a function in a different way is a good thing.
Proof
If we substitute from $\displaystyle B(p,q) = \int_{0}^{1} t^{p-1} (1-t)^{q-1} dt$ to $t = \sin^2 \theta$, $$ B(p,q) = \int_{0}^{{\pi} \over {2}} \left( \sin^2 \theta \right)^{p-1} \left( 1 - \sin^2 \theta \right) ^{q-1} 2 \sin \theta \cos \theta d \theta $$ since $1 - \sin^2 \theta = \cos ^2 \theta$, $$ B(p,q) = 2 \int_{0}^{{\pi} \over {2}} \left( \sin \theta \right)^{2p-1} \left( \cos \theta \right) ^{2q-1} d \theta $$
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Corollary
In particular, if we substitute again with $\sin \theta = t$, we obtain a lemma to derive Legendre’s duplication formula. $$ B(p,q) = 2 \int_{0}^{1} t^{2p-1} \left( 1 - t^2 \right)^{q-1} dt $$