1. **State the problem:** We need to evaluate the integral $$\int 45 \sin^3(x) \cos^2(x) \, dx.$$\n\n2. **Rewrite the integral:** Use the identity $$\sin^3(x) = \sin(x) \cdot \sin^2(x) = \sin(x)(1 - \cos^2(x))$$ to express the integral in terms of sine and cosine powers.\n\n3. **Substitute:** The integral becomes $$\int 45 \sin(x) (1 - \cos^2(x)) \cos^2(x) \, dx = \int 45 \sin(x) (\cos^2(x) - \cos^4(x)) \, dx.$$\n\n4. **Use substitution:** Let $$u = \cos(x)$$ so that $$du = -\sin(x) dx$$ or $$-du = \sin(x) dx.$$\n\n5. **Rewrite the integral in terms of $$u$$:** $$\int 45 \sin(x) (u^2 - u^4) \, dx = 45 \int (u^2 - u^4) \sin(x) dx = 45 \int (u^2 - u^4)(-du) = -45 \int (u^2 - u^4) du.$$\n\n6. **Integrate:** $$-45 \int (u^2 - u^4) du = -45 \left( \int u^2 du - \int u^4 du \right) = -45 \left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C.$$\n\n7. **Simplify:** $$-45 \left( \frac{u^3}{3} - \frac{u^5}{5} \right) = -45 \left( \frac{5u^3 - 3u^5}{15} \right) = -3 (5u^3 - 3u^5) + C = -15 u^3 + 9 u^5 + C.$$\n\n8. **Back-substitute $$u = \cos(x)$$:** $$\boxed{-15 \cos^3(x) + 9 \cos^5(x) + C}.$$