The integral you've provided is:

$\int \frac{x^3 \, dx}{\sqrt{9 - x^2}}$

To solve this integral, we can use trigonometric substitution. Here's the step-by-step process:

Step 1: Trigonometric Substitution

Let: $x = 3\sin\theta$ Then, $dx = 3\cos\theta \, d\theta$ and $\sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = 3\cos\theta$.

Step 2: Substitute into the Integral

Substituting these into the integral, we get: $\int \frac{(3\sin\theta)^3 \cdot 3\cos\theta \, d\theta}{3\cos\theta}$ This simplifies to: $27 \int \sin^3\theta \, d\theta$

Step 3: Simplify the Trigonometric Integral

We can use the identity $\sin^3\theta = \sin\theta(1 - \cos^2\theta)$: $27 \int \sin\theta (1 - \cos^2\theta) \, d\theta$ This can be split into two integrals: $27 \left( \int \sin\theta \, d\theta - \int \sin\theta \cos^2\theta \, d\theta \right)$

Step 4: Solve Each Integral

1. The first integral is straightforward: $\int \sin\theta \, d\theta = -\cos\theta$
2. For the second integral, we can use the substitution $u = \cos\theta$, $du = -\sin\theta \, d\theta$: $\int \sin\theta \cos^2\theta \, d\theta = -\int u^2 \, du = -\frac{u^3}{3} = -\frac{\cos^3\theta}{3}$

Step 5: Combine Results and Back-Substitute

Putting it all together: $27 \left(-\cos\theta + \frac{\cos^3\theta}{3}\right) + C$ Substitute back $\theta$ using $\sin\theta = \frac{x}{3}$ and $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{x^2}{9}}$: $-27\sqrt{1 - \frac{x^2}{9}} + \frac{9}{3} \left(1 - \frac{x^2}{9}\right)^{\frac{3}{2}} + C$

The final solution simplifies to: $\boxed{-27\sqrt{1 - \frac{x^2}{9}} + 9\left(1 - \frac{x^2}{9}\right)^{\frac{3}{2}} + C}$

Would you like further details or have any questions? Here are some related questions you might find interesting:

1. How does trigonometric substitution work for different types of integrals?
2. What are other common substitutions used in integrals involving square roots?
3. How does the Pythagorean identity help in solving trigonometric integrals?
4. What is the significance of the constant of integration, $C$, in indefinite integrals?
5. How can we verify the solution of an integral?

Tip: When solving integrals, always check if trigonometric identities or substitutions can simplify the problem significantly.