1. **State the problem:** We need to evaluate the integral $$\int \sec^3(3x) \tan(3x) \, dx.$$\n\n2. **Recall the formula and rules:** The integral involves powers of secant and tangent functions. A useful substitution is to let $$u = \sec(3x)$$ because the derivative of $$\sec(3x)$$ involves $$\sec(3x) \tan(3x)$$.\n\n3. **Substitution:** Let $$u = \sec(3x)$$. Then, $$\frac{du}{dx} = 3 \sec(3x) \tan(3x)$$, so $$du = 3 \sec(3x) \tan(3x) \, dx$$.\n\n4. **Rewrite the integral:** We have $$\int \sec^3(3x) \tan(3x) \, dx = \int u^3 \cdot \frac{du}{3u} = \frac{1}{3} \int u^2 \, du$$ because $$\sec^3(3x) \tan(3x) \, dx = u^3 \tan(3x) \, dx$$ and $$du = 3u \tan(3x) \, dx$$ so $$\tan(3x) \, dx = \frac{du}{3u}$$.\n\n5. **Simplify the integral:** $$\frac{1}{3} \int u^2 \, du = \frac{1}{3} \cdot \frac{u^3}{3} + C = \frac{u^3}{9} + C.$$\n\n6. **Back-substitute:** Replace $$u$$ with $$\sec(3x)$$ to get the final answer: $$\frac{\sec^3(3x)}{9} + C.$$