We need to evaluate the integral:

$I = \int_0^{2\sqrt{2}} \frac{-9x^2}{\sqrt{1 - x^2}} \, dx$

Step 1: Substitution to simplify the integral

Notice that the integrand contains $\sqrt{1 - x^2}$, which suggests using a trigonometric substitution. A common substitution for this form is:

$x = \sin(\theta), \quad dx = \cos(\theta) \, d\theta$

This simplifies $\sqrt{1 - x^2}$ because:

$\sqrt{1 - x^2} = \sqrt{1 - \sin^2(\theta)} = \cos(\theta)$

Also, when $x = 0$, $\theta = 0$, and when $x = 2\sqrt{2}$, the substitution is outside the range of $\sin(\theta)$, which is only valid for $x$ in $[-1, 1]$. This substitution will not be appropriate because the limits of integration exceed the valid range of $\sin(\theta)$.

Step 2: Rethinking the problem

Since the limits of integration go beyond $[-1, 1]$, a different approach or error-checking for the limits might be required, or perhaps a more appropriate substitution or transformation is needed.

Would you like me to reconsider the limits or explore an alternative method, such as substitution or numerical integration?