1. **State the problem:** We need to evaluate the integral $$\int 2x \sin x \, dx$$. 2. **Formula and rules:** Use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ Choose: $$u = 2x \implies du = 2 \, dx$$ $$dv = \sin x \, dx \implies v = -\cos x$$ 3. **Apply integration by parts:** $$\int 2x \sin x \, dx = 2x(-\cos x) - \int -\cos x (2) \, dx = -2x \cos x + 2 \int \cos x \, dx$$ 4. **Integrate remaining integral:** $$2 \int \cos x \, dx = 2 \sin x + C$$ 5. **Combine results:** $$\int 2x \sin x \, dx = -2x \cos x + 2 \sin x + C$$ **Final answer:** $$\boxed{-2x \cos x + 2 \sin x + C}$$