1. **State the problem:** Evaluate the definite integral $$\int_0^{\frac{\pi}{2}} \sin(2x) \, dx$$. 2. **Recall the formula:** The integral of $$\sin(ax)$$ with respect to $$x$$ is $$-\frac{1}{a} \cos(ax) + C$$, where $$a$$ is a constant. 3. **Apply the formula:** Here, $$a = 2$$, so $$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C$$. 4. **Evaluate the definite integral:** $$\int_0^{\frac{\pi}{2}} \sin(2x) \, dx = \left[-\frac{1}{2} \cos(2x)\right]_0^{\frac{\pi}{2}} = -\frac{1}{2} \cos(\pi) + \frac{1}{2} \cos(0)$$. 5. **Calculate the cosine values:** $$\cos(\pi) = -1$$ and $$\cos(0) = 1$$. 6. **Substitute and simplify:** $$-\frac{1}{2} (-1) + \frac{1}{2} (1) = \frac{1}{2} + \frac{1}{2} = 1$$. **Final answer:** $$\int_0^{\frac{\pi}{2}} \sin(2x) \, dx = 1$$.