Prove $\displaystyle B(p,q)=\int_{0}^{1}x^{p-1}(1-x)^{q-1}dx=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)}$
$\displaystyle B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q) }$....[1]
Start with $\displaystyle {\Gamma}(p)=\int_{0}^{\infty}t^{p-1}e^{-t}dt$
Now, put $\displaystyle t=y^{2}$, and we get:
$\displaystyle {\Gamma}(p)=2\int_{0}^{\infty}y^{2p-1}e^{-y^{2}}dy$....[2]
Similarly, (the dummy integration variable can be any letter):
$\displaystyle {\Gamma}(p)=2\int_{0}^{\infty}x^{2q-1}e^{-x^{2}}dx$
Next, we multiply these 2 equations and change to polar coordinates:
$\displaystyle {\Gamma}(p){\Gamma}(q)=4\int_{0}^{\infty}\int_{0}^ {\infty}x^{2q-1}y^{2p-1}e^{-(x^{2}+y^{2})}dxdy$
$\displaystyle =4\int_{0}^{\infty}\int_{0}^{\frac{\pi}{2}}(rcos{\ theta})^{2q-1}(rsin{\theta})^{2p-1}e^{-r^{2}}rdrd{\theta}$
$\displaystyle =4\int_{0}^{\infty}r^{2p+2q-1}e^{-r^{2}}dr\int_{0}^{\frac{\pi}{2}}(cos{\theta})^{2q-1}(sin{\theta})^{2p-1}d{\theta}$....[3]
The r integral in [3] is $\displaystyle \frac{1}{2}{\Gamma}(p+q)$ by [2].
The $\displaystyle {\theta}$ in [3] is $\displaystyle \frac{1}{2}B(p,q)$ by
$\displaystyle B(p,q)=2\int_{0}^{\frac{\pi}{2}}(sin{\theta})^{2p-1}(cos{\theta})^{2q-1}d{\theta}$
Then, $\displaystyle {\Gamma}(p){\Gamma}(q)=4\cdot \frac{1}{2}{\Gamma}(p+q)\cdot \frac{1}{2}B(p,q)$ and [1] follows.