By Euler's integral of the first kind in terms of Gamma functions:
$\int_0^{\pi/2} \sin^p x \cos^q x\,dx = \frac{\Gamma\left(\frac{p+1}{2}\right)\Gamma\left(\frac{q+1}{2}\right)}{2\Gamma\left(\frac{p+q+2}{2}\right)}$.
Setting $p = n$ and $q = 0$:
$\int_0^{\pi/2} \sin^n x\,dx = \frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{n+2}{2}\right)}$.
Since $\Gamma(1/2) = \sqrt{\pi}$, this simplifies to:
$\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} \frac{\sqrt{\pi}}{2}$.

Detailed analysis of options:

• $\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\frac{\sqrt{\pi}}{2}$ is the exact formula.
• Has $n/2$ and $(n+1)/2$ instead of $(n+1)/2$ and $(n+2)/2$.
• Inverts the ratio of Gamma arguments.
• Uses integer arguments $n$ and $n+1$.