The value of the integral $$\int_0^{\pi} |\sin 2x| dx$$ is ______.

We need to evaluate $$\int_0^{\pi} |\sin 2x| \, dx$$.

The function $$\sin 2x$$ changes sign at $$x = \frac{\pi}{2}$$. Specifically, $$\sin 2x \geq 0$$ for $$x \in [0, \frac{\pi}{2}]$$ and $$\sin 2x \leq 0$$ for $$x \in [\frac{\pi}{2}, \pi]$$.

So $$\int_0^{\pi} |\sin 2x| \, dx = \int_0^{\pi/2} \sin 2x \, dx + \int_{\pi/2}^{\pi} (-\sin 2x) \, dx$$.

For the first integral: $$\int_0^{\pi/2} \sin 2x \, dx = \left[-\frac{\cos 2x}{2}\right]_0^{\pi/2} = -\frac{\cos \pi}{2} + \frac{\cos 0}{2} = \frac{1}{2} + \frac{1}{2} = 1$$.

For the second integral: $$\int_{\pi/2}^{\pi} (-\sin 2x) \, dx = \left[\frac{\cos 2x}{2}\right]_{\pi/2}^{\pi} = \frac{\cos 2\pi}{2} - \frac{\cos \pi}{2} = \frac{1}{2} + \frac{1}{2} = 1$$.

Therefore $$\int_0^{\pi} |\sin 2x| \, dx = 1 + 1 = 2$$.