# Representing an Integral as gamma function

I'm trying to prove: $$\int_0^{\pi/2} (\cos \theta)^{2m-1}(\sin \theta)^{2n-1}d\theta =\dfrac {\Gamma (m)\Gamma (n)}{2\Gamma (m+n)}$$ Could you help me do that?

• what do you know on the subject ? what did you try? – G Cab Nov 17 '16 at 18:57
• I tried to convert polar coordinates to catesian and i got some result but I couldn't prove that exactly! – H.H Nov 17 '16 at 19:02

It's Beta function. Make a change of variable. Take $sin\theta = x$, then, $cos\theta = \sqrt{1-x^{2}}$. And the derivative $d\theta = \frac{1}{\sqrt{1-x^{2}}}$, and $\frac{\pi}{2}$ will become one. Than you will have form of $\int_{0}^{1}(1-x^{2})^{m-1}x^{2n-1}dx$. And make one more change of variable. $x^2 := t$. Than you will have $dx = \frac{1}{2}t^{-\frac{1}{2}}dt$. And you will have $$\int_{0}^{1}\frac{1}{2}t^{-\frac{1}{2}} (1-t)^{m-1}t^{n-\frac{1}{2}}dt=\frac{1}{2} \int_{0}^{1}(1-t)^{m-1}t^{n-1}dt= \frac{1}{2}B(m,n)= \frac{\Gamma(n)\Gamma(m)}{2\Gamma(n+m)}$$.
• In another way: \eqalign{ & {\rm B}(x,y) = 2\int_{\theta \, = \,0}^{\pi /2} {\cos (\theta )^{\,2\,x - 1} \sin (\theta )^{\,2\,y - 1} d\theta } = \cr & = 2\int_{\theta \, = \,0}^{\pi /2} {\cos (\theta )^{\,2\,x - 2} \sin (\theta )^{\,2\,y - 2} \cos (\theta )\sin (\theta )d\theta } = \cr & = \int_{\theta \, = \,0}^{\pi /2} {\cos (\theta )^{\,2\,x - 2} \sin (\theta )^{\,2\,y - 2} d\left( {\sin (\theta )^{\,2\,} } \right)} = \cr & = \int_{t\, = \,0}^{\,1} {\left( {1 - t} \right)^{\,x - 1} t^{\,\,y - 1} dt} \cr} – G Cab Nov 17 '16 at 23:18