Though a CS student, my main hobby is mathematics, and was wondering about the primitives of rational powers of $\sin x$.
I solved and found the primitive of the square root of $\sin x$ via a series and proceeded to use the same method for $3/2$, then $5/2$ and mostly gave the same results differing in a few key aspects regarding the powers of $\sin x$ in the series.
In all cases of positive integer $k$, your integral is equivalent to $$I_k(x)=\int_0^x\sin^{k/2}(t)dt$$ So let's focus on a different integral for a second: $$G(x,a,b)=\int_0^x\sin^a(t)\cos^b(t)dt$$ To evaluate this, we make the substitution $u=\sin^2(t)$, giving $dt=\frac12u^{-1/2}(1-u)^{-1/2}du$, which gives $$G(x,a,b)=\int_0^{\sin^2(x)}u^{a/2}(1-u)^{b/2}\frac12u^{-1/2}(1-u)^{-1/2}du$$ $$G(x,a,b)=\frac12\int_0^{\sin^2(x)}u^{\frac{a-1}2}(1-u)^{\frac{b-1}2}du$$ $$G(x,a,b)=\frac12\int_0^{\sin^2(x)}u^{\frac{a+1}2-1}(1-u)^{\frac{b+1}2-1}du$$ Next we recall the definition of the incomplete beta function: $$B(x;a,b)=\int_0^xt^{a-1}(1-t)^{b-1}dt$$ Which gives our generalized integral: $$G(x,a,b)=\frac12 B\bigg(\sin^2(x);\frac{a+1}2,\frac{b+1}2\bigg)$$ Then back to the integral in question: $$I_k(x)=G\bigg(x,\frac k2,0\bigg)$$ $$I_k(x)=\frac12 B\bigg(\sin^2(x);\frac{k+2}4,\frac12\bigg)$$ A Special Value:
Note that $$B(1;a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ Where $\Gamma(s)$ is the Gamma function.
Therefore, $$I_k\bigg(\frac\pi2\bigg)=\frac12 B\bigg(1;\frac{k+2}4,\frac12\bigg)$$ $$I_k\bigg(\frac\pi2\bigg)=\frac{\Gamma(\frac{k+2}4)\Gamma(\frac12)}{2\Gamma(\frac{k+4}4)}$$ And from $\Gamma(\frac12)=\sqrt{\pi}$, $$I_k\bigg(\frac\pi2\bigg)=\frac{\sqrt{\pi}\,\Gamma(\frac{k+2}4)}{2\Gamma(\frac{k+4}4)}$$ Which doesn't really get anymore simplified than that.