1. The problem is to find the integral of $\sin 2x$ with respect to $x$, i.e., $\int \sin 2x \, dx$. 2. Recall the formula for integrating sine functions: $\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C$, where $a$ is a constant and $C$ is the constant of integration. 3. Here, $a = 2$, so applying the formula: $$\int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C$$ 4. This means the antiderivative of $\sin 2x$ is $-\frac{1}{2} \cos 2x$ plus the constant of integration. 5. Therefore, the final answer is: $$\boxed{-\frac{1}{2} \cos 2x + C}$$