1. **State the problem:** Evaluate the integral $$\int 4 \sin x \cos x \, dx$$. 2. **Recall the formula:** Use the double-angle identity for sine: $$\sin(2x) = 2 \sin x \cos x$$. 3. **Rewrite the integrand:** $$4 \sin x \cos x = 2 \cdot 2 \sin x \cos x = 2 \sin(2x)$$. 4. **Substitute into the integral:** $$\int 4 \sin x \cos x \, dx = \int 2 \sin(2x) \, dx$$. 5. **Integrate:** Use the integral formula $$\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C$$. So, $$\int 2 \sin(2x) \, dx = 2 \int \sin(2x) \, dx = 2 \left(-\frac{1}{2} \cos(2x)\right) + C = -\cos(2x) + C$$. 6. **Final answer:** $$\boxed{-\cos(2x) + C}$$