1. **State the problem:** Evaluate the integral $$\int_0^{\frac{\pi}{6}} (1 - \cos 3x) \sin 3x \, dx$$. 2. **Recall the formula and rules:** We will use substitution and trigonometric identities. The integral involves products of sine and cosine functions. 3. **Rewrite the integral:** $$\int_0^{\frac{\pi}{6}} (1 - \cos 3x) \sin 3x \, dx = \int_0^{\frac{\pi}{6}} \sin 3x \, dx - \int_0^{\frac{\pi}{6}} \cos 3x \sin 3x \, dx$$ 4. **Evaluate the first integral:** Use substitution $u = 3x$, so $du = 3 dx$, $dx = \frac{du}{3}$. $$\int_0^{\frac{\pi}{6}} \sin 3x \, dx = \int_0^{\frac{\pi}{2}} \sin u \frac{du}{3} = \frac{1}{3} \int_0^{\frac{\pi}{2}} \sin u \, du$$ $$= \frac{1}{3} [-\cos u]_0^{\frac{\pi}{2}} = \frac{1}{3} [-\cos \frac{\pi}{2} + \cos 0] = \frac{1}{3} [0 + 1] = \frac{1}{3}$$ 5. **Evaluate the second integral:** Use substitution $v = \sin 3x$, then $dv = 3 \cos 3x \, dx$, so $\cos 3x \, dx = \frac{dv}{3}$. $$\int_0^{\frac{\pi}{6}} \cos 3x \sin 3x \, dx = \int_{v=\sin 0}^{v=\sin \frac{\pi}{2}} v \frac{dv}{3} = \frac{1}{3} \int_0^1 v \, dv = \frac{1}{3} \left[ \frac{v^2}{2} \right]_0^1 = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$$ 6. **Combine results:** $$\int_0^{\frac{\pi}{6}} (1 - \cos 3x) \sin 3x \, dx = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$$ **Final answer:** $$\boxed{\frac{1}{6}}$$