We need to evaluate the integral $I = \int \frac{x^3}{\sqrt{4-x^2}} \, dx$.

1. Problem Description

We need to evaluate the integral $I = \int \frac{x^3}{\sqrt{4-x^2}} \, dx$.

2. Solution Steps

We will use substitution.

Let $u = 4 - x^2$. Then $du = -2x \, dx$, so $x \, dx = -\frac{1}{2} \, du$.

Also, $x^2 = 4 - u$.

Then the integral becomes

$I = \int \frac{x^2 \cdot x}{\sqrt{4-x^2}} \, dx = \int \frac{(4-u)}{\sqrt{u}} (-\frac{1}{2}) \, du = -\frac{1}{2} \int \frac{4-u}{\sqrt{u}} \, du = -\frac{1}{2} \int (4u^{-1/2} - u^{1/2}) \, du$

Now, we can integrate term by term:

$I = -\frac{1}{2} \left[ 4 \int u^{-1/2} \, du - \int u^{1/2} \, du \right]$

Recall the power rule for integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, for $n \ne -1$.

$I = -\frac{1}{2} \left[ 4 \cdot \frac{u^{1/2}}{1/2} - \frac{u^{3/2}}{3/2} \right] + C$

$I = -\frac{1}{2} \left[ 8u^{1/2} - \frac{2}{3}u^{3/2} \right] + C$

$I = -4u^{1/2} + \frac{1}{3}u^{3/2} + C$

Substitute back $u = 4-x^2$:

$I = -4\sqrt{4-x^2} + \frac{1}{3}(4-x^2)^{3/2} + C$

$I = \sqrt{4-x^2} \left[ -4 + \frac{1}{3}(4-x^2) \right] + C$

$I = \sqrt{4-x^2} \left[ -4 + \frac{4}{3} - \frac{x^2}{3} \right] + C$

$I = \sqrt{4-x^2} \left[ \frac{-12+4}{3} - \frac{x^2}{3} \right] + C$

$I = \sqrt{4-x^2} \left[ -\frac{8}{3} - \frac{x^2}{3} \right] + C$

$I = -\frac{1}{3}(8+x^2)\sqrt{4-x^2} + C$

3. Final Answer

$I = -\frac{1}{3}(x^2+8)\sqrt{4-x^2} + C$