Denote $$ I_k=\int\sin^k xdx\qquad J_k=\int_0^\pi \sin^k xdx $$ Integration by parts gives $$ \begin{align} I_k &=-\sin^{k-1}x\cos x-\int \cos x(k-1)\sin^{k-2}x\cos xdx\\ &=-\sin^{k-1}x\cos x+(k-1)\int \sin^{k-2}x(1-\sin^2 x) xdx\\ &=-\sin^{k-1}x\cos x+(k-1)I_{k-2}(x)-(k-1)I_k(x)\\ \end{align} $$ So we derive the recurrence formula $$ I_k=-\frac{\sin^{k-1}\cos x}{k}+\frac{k-1}{k} I_{k-2}(x) $$ Hence the desired integral is given by the recurrence formula $$ J_k=\frac{k-1}{k}J_{k-2} $$ Note that $J_1=2$, $J_2=\pi/2$. From here you can easily verify that $$ J_{2k}=\frac{(2k-1)(2k-3)\cdot\ldots\cdot 3}{2k(2k-2)\cdot\ldots\cdot 4}\cdot 2\\ J_{2k+1}=\frac{2k(2k-2)\cdot\ldots\cdot 2}{(2k+1)(2k-1)\cdot\ldots\cdot 3}\cdot \frac{\pi}{2} $$ From here you can verify that $$ J_n=\sqrt\pi\frac{\Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})} $$

If you use the tangent half-angle substitution, the antiderivative involves the hypergeometric 2F1 function. Taking the values at the limits gives what you report.
If I properly remember, Sin[x]^n can be represented as a linear combination of sigle powers of Cos (if n is even) and as a linear combination of sigle powers of Sin (if n is odd). This probably could help.