Use Beta Function Identity to Prove the Equality

Question

How to prove the equality $${ {2}^{a+b}}\ B\left( {a+b+1\over2} ,{a+b+1\over2}\right)=B\left( {1\over2} ,{a+b+1\over2}\right), $$ where $B(x,y)$ represents the beta function.

Answer 1

$$B(p,q)=2\int_{0}^{\pi/2} \sin^{2p-1}x \cos^{2q-1} x dx$$ $\Rightarrow$ $$B(p,p) = 2^{2-2p}\int_{0}^{\pi/2} (\sin 2x)^{2p-1} dx= 2^{1-2p} \int_{0}^{\pi} (\sin y)^{2p-1} dy $$ $$= 2^{1-2p} \, 2 \int _{0}^{\pi/2} (\sin y)^{2p-1} y \cos^0 y~ dy =2^{1-2p} ~B(p,\frac{1}{2}) =2^{1-2p}~ B(\frac{1}{2}, p).$$ In the case here $p=(a+b+1)/2.$