Consider the integral $$I(a,b)=\int_0^{\pi/2}\sin^at\ \cos^bt\ dt$$ As I have shown in numerous answers, $I$ has a close relationship with the beta function, namely $$I(a,b)=\frac12B\bigg(\frac{a+1}2,\frac{b+1}2\bigg)$$ Which of course lends itself to a representation via the gamma function $$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ And similarly, $$\int_0^x\sin^at\ \cos^bt\ dt=\frac12B\bigg(\sin^2x;\frac{a+1}2,\frac{b+1}2\bigg)$$ Where $B(x;a,b)$ is the Incomplete Beta function.
But here's my question:
Let $$G_{ab}(x)=\frac12B\bigg(\sin^2x;\frac{a+1}2,\frac{b+1}2\bigg)$$ Assuming $a,b>-1$ are held constant, what is the relationship between $G_{ab}(\frac{\pi}2)$ and $G_{ab}(x)$ at other values of $x\geq0$. I don't expect that there is a closed form expression for all $x$, but I'd expect there'd be one for $$x\in\bigg\{\frac{\pi}2(2n+1):n\in\Bbb N\bigg\}$$ As $\sin^2x=1$ for such $x$.
My instinct tells me that for such $x$, $$G_{ab}(x)=I(a,b)$$ But I just don't know how to prove it. Could I have some help?