Is there a general pattern for all integrals of the form $\sin^{k/2}x$ where $k$ is an integer? 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. 
 A: 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.
