Csdrhrt

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
$$I_k(x)=int_0^xsin^{k/2}(t)dt$$
So let's focus on a different integral for a second:
$$G(x,a,b)=int_0^xsin^a(t)cos^b(t)dt$$
To evaluate this, we make the substitution $u=sin^2(t)$, giving $dt=frac12u^{-1/2}(1-u)^{-1/2}du$, which gives
$$G(x,a,b)=int_0^{sin^2(x)}u^{a/2}(1-u)^{b/2}frac12u^{-1/2}(1-u)^{-1/2}du$$
$$G(x,a,b)=frac12int_0^{sin^2(x)}u^{frac{a-1}2}(1-u)^{frac{b-1}2}du$$
$$G(x,a,b)=frac12int_0^{sin^2(x)}u^{frac{a+1}2-1}(1-u)^{frac{b+1}2-1}du$$
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:
$$G(x,a,b)=frac12 Bbigg(sin^2(x);frac{a+1}2,frac{b+1}2bigg)$$
Then back to the integral in question:
$$I_k(x)=Gbigg(x,frac k2,0bigg)$$
$$I_k(x)=frac12 Bbigg(sin^2(x);frac{k+2}4,frac12bigg)$$
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,
$$I_kbigg(fracpi2bigg)=frac12 Bbigg(1;frac{k+2}4,frac12bigg)$$
$$I_kbigg(fracpi2bigg)=frac{Gamma(frac{k+2}4)Gamma(frac12)}{2Gamma(frac{k+4}4)}$$
And from $Gamma(frac12)=sqrt{pi}$,
$$I_kbigg(fracpi2bigg)=frac{sqrt{pi},Gamma(frac{k+2}4)}{2Gamma(frac{k+4}4)}$$
Which doesn't really get anymore simplified than that.