So, I am working on $$s(\alpha)=\int_{0}^{\pi/2}\sin^{\alpha}(t)dt$$ Looking for a general form. Although I am not (really) asking you to evaluate the integral, I have some questions about my methods. Here's what I've done so far.
For all $t\in[0,\frac{\pi}{2}]$, $$I(t,\alpha)=\int\sin^{\alpha}(t)dt\\\therefore I(t,a)=(2i)^{-\alpha}\int\big(e^{it}-e^{-it}\big)^{\alpha}dt$$ Then, I note that for all $x\in(0,2)$, and any $\beta$, $$x^\beta=\,_1F_0(-\beta;;1-x)$$ My plan is to turn $\big(e^{it}-e^{-it}\big)^{\alpha}$ into $$_1F_0(-\alpha;;1-e^{it}+e^{-it})$$ expand the series, turn $1-e^{it}+e^{-it}$ into some trinomial expansion using the trinomial formula, then integrate term by term. But I am confused on whether or not this will work. But based on what I have previously stated, $_1F_0(-\alpha;;1-e^{it}+e^{-it})$ is only defined for $e^{-it}-e^{it}\in(0,2)$ which makes me worry that I won't be able carry out my plan, because $[0,\frac{\pi}{2}]\not\subset(0,2)$.
Ideally, you would reassure me that I am on the right path to a solution, but not everything goes my way. So, if I'm doing something wrong, or I can't get to a solution the way I want to, please inform me and show me how to do it correctly.
This can actually be expressed in closed form in terms of the beta function as $$ \int_0^\frac\pi2 \sin^\alpha (t) dt = \frac12B\left(\frac{\alpha+1}2,\frac12\right) $$ which can be rewritten in terms of the Gamma function as $$ \int_0^\frac\pi2 \sin^\alpha (t) dt = \frac12B\left(\frac{\alpha+1}2,\frac12\right) = \frac{\Gamma(\frac{\alpha+1}2)\Gamma(\frac12)}{2\Gamma(\frac{\alpha}2 + 1)} =\frac{\sqrt{\pi}}2\frac{\Gamma(\frac{\alpha+1}2)}{\Gamma(\frac{\alpha}2 + 1)} $$ This can be seen by making the substitution $u = \sin^2(t)$, which gives \begin{eqnarray} t &=& \arcsin{\sqrt{u}}\\ dt &=& \frac{1}{2\sqrt{u(1-u)}} \end{eqnarray} Putting this into the original integral, obtain $$ \int_0^\frac\pi2 \sin^\alpha (t) dt = \int_0^1\frac{u^{\frac{\alpha - 1}2}}{2\sqrt{1-u}} du = \frac12B\left(\frac{\alpha + 1}2,\frac12\right) $$ as claimed above.