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

1. Problem Description

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

2. Solution Steps

We can solve this integral using substitution.

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

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

We can rewrite the integral as:

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

$I = -\frac{1}{2} \int (4u^{-1/2} - u^{1/2}) du = -\frac{1}{2} \left(4 \int u^{-1/2} du - \int u^{1/2} du \right)$

Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:

$\int u^{-1/2} du = \frac{u^{1/2}}{1/2} = 2u^{1/2}$

$\int u^{1/2} du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$

Therefore,

$I = -\frac{1}{2} \left(4 (2u^{1/2}) - \frac{2}{3} u^{3/2} \right) + C = -\frac{1}{2} \left(8 u^{1/2} - \frac{2}{3} u^{3/2} \right) + C$

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

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

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

We can also write this as:

$I = \sqrt{4-x^2} \left(\frac{1}{3}(4-x^2) - 4\right) + C = \sqrt{4-x^2} \left(\frac{4}{3} - \frac{x^2}{3} - \frac{12}{3}\right) + C$

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

3. Final Answer

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