Integration of powers of the $\sin x$

The answer provided by Leo is the first one that comes to mind, but here is one starting directly from the definition of $\Gamma(s)$.

From the definition of Gamma:

Consider

$$\Gamma(s)\Gamma(z)=\int_{0}^{\infty}\int_{0}^{\infty}t^{s-1}u^{z-1}e^{-(t+u)}dtdu.$$

Let $t=x^{2}$, $u=y^{2}$. Then we have

$$\Gamma(s)\Gamma(z)=4\int_{0}^{\infty}\int_{0}^{\infty}x^{2s-1}y^{2z-1}e^{-(x^{2}+y^{2})}dtdu.$$

Change to polar coordinates and set $y=r\sin\theta$, $x=r\cos\theta$, to get

$$\Gamma(s)\Gamma(z)=4\left(\int_{0}^{\pi/2}\cos^{2s-1}\theta\sin^{2z-1}\theta d\theta\right)\left(\int_{0}^{\infty}r^{2s+2z-1}e^{-r^{2}}dr\right).$$

Letting $\eta=r^{2}$ we get

$$2\int_{0}^{\infty}r^{2s+2z-1}e^{-r^{2}}dr=\int_{0}^{\infty}\eta^{s+z-1}e^{-\eta}d\eta=\Gamma(s+z).$$

Hence

$$\frac{\Gamma(s)\Gamma(z)}{\Gamma(s+z)}=2\left(\int_{0}^{\pi/2}\cos^{2s-1}\theta \sin^{2z-1}\theta d\theta\right).$$

Setting $s=\frac{1}{2}$ and $z=\frac{x+1}{2}$ then yields your identity.

Hope that helps,

Just, following Theo's hint $$ \begin{align*} \int_{0}^{\frac{\pi}{2}}{(\sin\psi)^x}d\psi&= \int_{0}^{\frac{\pi}{2}}{(\sin\psi)^{2\cdot \frac{1}{2}(x+1)-1}(\cos\psi)^{2\cdot \frac{1}{2}-1}}d\psi\\ &=\frac{1}{2}B\left( \frac{x+1}{2},\frac{1}{2} \right)\\ &= \frac{1}{2}\cdot \frac{\Gamma\left(\frac{x+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left( \frac{x}{2}+1 \right)}\\ &=\frac{\sqrt{\pi}\Gamma\left(\frac{x+1}{2}\right)}{2\Gamma\left( \frac{x}{2}+1 \right)}. \end{align*}$$