1. **State the problem:** Find the exact value of the integral $$\int_0^{\frac{\pi}{6}} x^2 \sin(2x) \, dx$$. 2. **Formula and method:** We will use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = x^2 \implies du = 2x \, dx$$ $$dv = \sin(2x) \, dx \implies v = -\frac{1}{2} \cos(2x)$$ 4. **Apply integration by parts:** $$\int x^2 \sin(2x) \, dx = -\frac{1}{2} x^2 \cos(2x) + \int x \cos(2x) \, dx$$ 5. **Second integration by parts for** $$\int x \cos(2x) \, dx$$: Let $$u = x \implies du = dx$$ $$dv = \cos(2x) \, dx \implies v = \frac{1}{2} \sin(2x)$$ 6. **Apply again:** $$\int x \cos(2x) \, dx = \frac{1}{2} x \sin(2x) - \int \frac{1}{2} \sin(2x) \, dx$$ 7. **Integrate remaining integral:** $$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)$$ So, $$\int x \cos(2x) \, dx = \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) + C$$ 8. **Combine results:** $$\int x^2 \sin(2x) \, dx = -\frac{1}{2} x^2 \cos(2x) + \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) + C$$ 9. **Evaluate definite integral from 0 to $$\frac{\pi}{6}$$:** $$\left[-\frac{1}{2} x^2 \cos(2x) + \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) \right]_0^{\frac{\pi}{6}}$$ 10. **Calculate at upper limit $$x=\frac{\pi}{6}$$:** $$-\frac{1}{2} \left(\frac{\pi}{6}\right)^2 \cos\left(\frac{\pi}{3}\right) + \frac{1}{2} \left(\frac{\pi}{6}\right) \sin\left(\frac{\pi}{3}\right) + \frac{1}{4} \cos\left(\frac{\pi}{3}\right)$$ Recall: $$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ So, $$-\frac{1}{2} \cdot \frac{\pi^2}{36} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + \frac{1}{4} \cdot \frac{1}{2} = -\frac{\pi^2}{144} + \frac{\pi \sqrt{3}}{24} + \frac{1}{8}$$ 11. **Calculate at lower limit $$x=0$$:** $$-\frac{1}{2} \cdot 0^2 \cdot \cos(0) + \frac{1}{2} \cdot 0 \cdot \sin(0) + \frac{1}{4} \cdot \cos(0) = \frac{1}{4}$$ 12. **Subtract lower from upper:** $$\left(-\frac{\pi^2}{144} + \frac{\pi \sqrt{3}}{24} + \frac{1}{8}\right) - \frac{1}{4} = -\frac{\pi^2}{144} + \frac{\pi \sqrt{3}}{24} - \frac{1}{8}$$ **Final answer:** $$\boxed{-\frac{\pi^2}{144} + \frac{\pi \sqrt{3}}{24} - \frac{1}{8}}$$