1. **State the problem:** Evaluate the integral $$\int \sec^2(6x) \tan^5(6x) \, dx$$. 2. **Recall the formula and substitution:** We know that $$\frac{d}{dx} \tan(6x) = 6 \sec^2(6x)$$. 3. **Use substitution:** Let $$u = \tan(6x)$$, then $$du = 6 \sec^2(6x) \, dx$$ or $$\frac{du}{6} = \sec^2(6x) \, dx$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int \sec^2(6x) \tan^5(6x) \, dx = \int u^5 \cdot \sec^2(6x) \, dx = \int u^5 \cdot \frac{du}{6} = \frac{1}{6} \int u^5 \, du$$. 5. **Integrate:** $$\frac{1}{6} \int u^5 \, du = \frac{1}{6} \cdot \frac{u^6}{6} = \frac{u^6}{36} + C$$. 6. **Back-substitute:** $$\frac{\tan^6(6x)}{36} + C$$. **Final answer:** $$\int \sec^2(6x) \tan^5(6x) \, dx = \frac{\tan^6(6x)}{36} + C$$.