The problem asks to evaluate the integral: $I = \int \frac{x^3}{\sqrt{4 - x^2}} dx$

1. Problem Description

The problem asks to evaluate the integral:

$I = \int \frac{x^3}{\sqrt{4 - x^2}} dx$

2. Solution Steps

First, we make the trigonometric substitution $x = 2 \sin \alpha$.

Then $dx = 2 \cos \alpha \, d\alpha$. Substituting these into the integral, we have:

$I = \int \frac{(2 \sin \alpha)^3}{\sqrt{4 - (2 \sin \alpha)^2}} (2 \cos \alpha) \, d\alpha = \int \frac{8 \sin^3 \alpha}{\sqrt{4 - 4 \sin^2 \alpha}} (2 \cos \alpha) \, d\alpha$

$I = \int \frac{16 \sin^3 \alpha \cos \alpha}{\sqrt{4(1 - \sin^2 \alpha)}} \, d\alpha = \int \frac{16 \sin^3 \alpha \cos \alpha}{2 \sqrt{\cos^2 \alpha}} \, d\alpha = \int \frac{16 \sin^3 \alpha \cos \alpha}{2 \cos \alpha} \, d\alpha$

$I = \int 8 \sin^3 \alpha \, d\alpha = 8 \int \sin^3 \alpha \, d\alpha$

We know that $\sin^3 \alpha = \sin \alpha (1 - \cos^2 \alpha) = \sin \alpha - \sin \alpha \cos^2 \alpha$. Thus,

$I = 8 \int (\sin \alpha - \sin \alpha \cos^2 \alpha) \, d\alpha = 8 \int \sin \alpha \, d\alpha - 8 \int \sin \alpha \cos^2 \alpha \, d\alpha$

Let $u = \cos \alpha$, then $du = - \sin \alpha \, d\alpha$. So the second integral becomes $\int \sin \alpha \cos^2 \alpha \, d\alpha = - \int u^2 du = - \frac{u^3}{3} = - \frac{\cos^3 \alpha}{3}$.

Therefore,

$I = 8(-\cos \alpha) - 8(-\frac{\cos^3 \alpha}{3}) + C = -8 \cos \alpha + \frac{8}{3} \cos^3 \alpha + C$

Since $x = 2 \sin \alpha$, we have $\sin \alpha = \frac{x}{2}$, and $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}$.

Substituting this back into the expression for $I$, we get:

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

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

$I = \sqrt{4 - x^2} (\frac{-12 + 4 - x^2}{3}) + C = \sqrt{4 - x^2} (\frac{-8 - x^2}{3}) + C = - \frac{1}{3} (8 + x^2) \sqrt{4 - x^2} + C$

3. Final Answer

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