1. **State the problem:** Evaluate the integral $$\int 5 \sin^3(3x) \cos^3(3x) \, dx$$. 2. **Recall the formula and rules:** We can use the identity for powers of sine and cosine and substitution. Note that $$\sin^3(\theta) = \sin(\theta) \cdot \sin^2(\theta)$$ and $$\sin^2(\theta) = 1 - \cos^2(\theta)$$. Similarly for cosine. 3. **Rewrite the integral:** $$\int 5 \sin^3(3x) \cos^3(3x) \, dx = \int 5 \sin(3x) (\sin^2(3x)) \cos^3(3x) \, dx = \int 5 \sin(3x) (1 - \cos^2(3x)) \cos^3(3x) \, dx$$ 4. **Substitute:** Let $$u = \cos(3x)$$, then $$du = -3 \sin(3x) dx$$ or $$\sin(3x) dx = -\frac{1}{3} du$$. 5. **Rewrite the integral in terms of $$u$$:** $$\int 5 \sin(3x) (1 - u^2) u^3 \, dx = 5 \int (1 - u^2) u^3 \sin(3x) dx = 5 \int (1 - u^2) u^3 \left(-\frac{1}{3} du\right) = -\frac{5}{3} \int (1 - u^2) u^3 du$$ 6. **Simplify the integrand:** $$(1 - u^2) u^3 = u^3 - u^5$$ 7. **Integrate:** $$-\frac{5}{3} \int (u^3 - u^5) du = -\frac{5}{3} \left( \frac{u^4}{4} - \frac{u^6}{6} \right) + C = -\frac{5}{3} \left( \frac{u^4}{4} - \frac{u^6}{6} \right) + C$$ 8. **Simplify constants:** $$= -\frac{5}{3} \cdot \frac{u^4}{4} + \frac{5}{3} \cdot \frac{u^6}{6} + C = -\frac{5 u^4}{12} + \frac{5 u^6}{18} + C$$ 9. **Substitute back $$u = \cos(3x)$$:** $$= -\frac{5}{12} \cos^4(3x) + \frac{5}{18} \cos^6(3x) + C$$ **Final answer:** $$\int 5 \sin^3(3x) \cos^3(3x) \, dx = -\frac{5}{12} \cos^4(3x) + \frac{5}{18} \cos^6(3x) + C$$