Show that $\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\sqrt{\frac{p+1}{2}}\sqrt{\frac{q+1}{2}}}{2\sqrt{\frac{p+q+2}{2}}},\; p,q > -1$ 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.


if $p=q=1$,  we get
$$\int_0^\frac {\pi}{2} \frac {\sin (2\theta)d\theta}{2} $$
$$=\frac {1}{4}[-\cos (2\theta)]_0^\frac {\pi}{2} $$
$$=\frac {1}{2} \neq \frac {1}{2\sqrt {2}} $$
Your formula doesn't seem to be correct.
