TITLE: Show that $\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\sqrt{\frac{p+1}{2}}\sqrt{\frac{q+1}{2}}}{2\sqrt{\frac{p+q+2}{2}}},\; p,q > -1$
QUESTION [5 upvotes]: Show that: $$\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\sqrt{\frac{p+1}{2}}\sqrt{\frac{q+1}{2}}}{2\sqrt{\frac{p+q+2}{2}}},\; p,q > -1$$
Here's the question.

REPLY [7 votes]: I think that the OP meant the Gamma function instead of square root. If so the relation is true, as from the Beta function we have:
$$\int_0^1 v^{y-1}(1-v)^{x-1} dv = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
Then substituting $v = \sin^2 \theta$ will provide you with the result.