1. **State the problem:** We need to evaluate the integral $$\int \frac{x^3}{\sqrt{9 - x^2}} \, dx$$. 2. **Recall the formula and substitution:** When dealing with integrals involving expressions like $\sqrt{a^2 - x^2}$, a common substitution is $x = a \sin \theta$ because $1 - \sin^2 \theta = \cos^2 \theta$ simplifies the square root. 3. **Apply substitution:** Let $$x = 3 \sin \theta$$, so $$dx = 3 \cos \theta \, d\theta$$. 4. **Rewrite the integral:** Substitute into the integral: $$\int \frac{(3 \sin \theta)^3}{\sqrt{9 - (3 \sin \theta)^2}} \cdot 3 \cos \theta \, d\theta = \int \frac{27 \sin^3 \theta}{\sqrt{9 - 9 \sin^2 \theta}} \cdot 3 \cos \theta \, d\theta$$ 5. **Simplify the square root:** $$\sqrt{9 - 9 \sin^2 \theta} = \sqrt{9(1 - \sin^2 \theta)} = \sqrt{9 \cos^2 \theta} = 3 \cos \theta$$ 6. **Simplify the integral:** $$\int \frac{27 \sin^3 \theta}{3 \cos \theta} \cdot 3 \cos \theta \, d\theta = \int 27 \sin^3 \theta \, d\theta$$ 7. **Cancel common factors:** $$\int \cancel{\frac{27 \sin^3 \theta}{3 \cos \theta}} \cdot \cancel{3 \cos \theta} \, d\theta = \int 27 \sin^3 \theta \, d\theta$$ 8. **Rewrite the integral:** $$27 \int \sin^3 \theta \, d\theta$$ 9. **Use the identity for $\sin^3 \theta$:** $$\sin^3 \theta = \sin \theta (1 - \cos^2 \theta)$$ 10. **Substitute and integrate:** $$27 \int \sin \theta (1 - \cos^2 \theta) \, d\theta$$ Let $$u = \cos \theta$$, so $$du = -\sin \theta \, d\theta$$, thus $$-du = \sin \theta \, d\theta$$. 11. **Rewrite integral in terms of $u$:** $$27 \int (1 - u^2)(-du) = -27 \int (1 - u^2) \, du = -27 \left( \int 1 \, du - \int u^2 \, du \right)$$ 12. **Integrate:** $$-27 \left( u - \frac{u^3}{3} \right) + C = -27 u + 9 u^3 + C$$ 13. **Back-substitute $u = \cos \theta$:** $$-27 \cos \theta + 9 \cos^3 \theta + C$$ 14. **Back-substitute $\theta$ to $x$:** Recall $x = 3 \sin \theta$, so $\sin \theta = \frac{x}{3}$ and $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{x^2}{9}} = \frac{\sqrt{9 - x^2}}{3}$. 15. **Final answer:** $$-27 \cdot \frac{\sqrt{9 - x^2}}{3} + 9 \left( \frac{\sqrt{9 - x^2}}{3} \right)^3 + C = -9 \sqrt{9 - x^2} + 9 \cdot \frac{(9 - x^2)^{3/2}}{27} + C = -9 \sqrt{9 - x^2} + \frac{1}{3} (9 - x^2)^{3/2} + C$$