1. **State the problem:** We need to evaluate the definite integral $$\int_0^{\frac{\pi}{6}} x \sin(2x) \, dx$$. 2. **Use integration by parts:** Let $$u = x$$ and $$dv = \sin(2x) dx$$. 3. **Compute derivatives and integrals:** - $$du = dx$$ - To find $$v$$, integrate $$dv$$: $$v = \int \sin(2x) dx = -\frac{1}{2} \cos(2x)$$. 4. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ So, $$\int_0^{\frac{\pi}{6}} x \sin(2x) dx = \left[-\frac{x}{2} \cos(2x)\right]_0^{\frac{\pi}{6}} + \frac{1}{2} \int_0^{\frac{\pi}{6}} \cos(2x) dx$$ 5. **Evaluate the remaining integral:** $$\int \cos(2x) dx = \frac{1}{2} \sin(2x)$$ So, $$\frac{1}{2} \int_0^{\frac{\pi}{6}} \cos(2x) dx = \frac{1}{2} \left[ \frac{1}{2} \sin(2x) \right]_0^{\frac{\pi}{6}} = \frac{1}{4} \left( \sin\left(\frac{\pi}{3}\right) - \sin(0) \right) = \frac{1}{4} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$$ 6. **Evaluate the boundary term:** $$-\frac{x}{2} \cos(2x) \Big|_0^{\frac{\pi}{6}} = -\frac{\pi}{12} \cos\left(\frac{\pi}{3}\right) + 0 = -\frac{\pi}{12} \times \frac{1}{2} = -\frac{\pi}{24}$$ 7. **Combine results:** $$\int_0^{\frac{\pi}{6}} x \sin(2x) dx = -\frac{\pi}{24} + \frac{\sqrt{3}}{8}$$ **Final answer:** $$\boxed{\frac{\sqrt{3}}{8} - \frac{\pi}{24}}$$