To solve the integral

$\int_0^{\frac{2}{3}} -6x^3 \sqrt{4 - 9x^2} \, dx$

we will follow these steps:

Step 1: Simplify the expression

We are given the integrand as $-6x^3 \sqrt{4 - 9x^2}$. This doesn't simplify easily, but we can try a trigonometric substitution to handle the square root term.

Step 2: Trigonometric substitution

The term inside the square root is of the form $a^2 - b^2x^2$, where $a = 2$ and $b = 3$. This suggests using the substitution:

$x = \frac{2}{3} \sin(\theta)$ which leads to:

$dx = \frac{2}{3} \cos(\theta) d\theta$

Now, substitute $x = \frac{2}{3} \sin(\theta)$ into the integrand:

$\sqrt{4 - 9x^2} = \sqrt{4 - 9 \left(\frac{4}{9} \sin^2(\theta)\right)} = \sqrt{4(1 - \sin^2(\theta))} = 2 \cos(\theta)$

Step 3: Apply the substitution to the entire integrand

The integrand becomes:

$-6 \left(\frac{2}{3} \sin(\theta)\right)^3 2 \cos(\theta) \cdot \frac{2}{3} \cos(\theta) d\theta$

Simplify the expression step by step.

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Here are some relative questions for further exploration:

1. How can we handle integrals involving square roots?
2. What is the typical process for applying trigonometric substitutions in integrals?
3. Can we generalize the substitution $x = \frac{a}{b} \sin(\theta)$ for other integrals?
4. How do you deal with integrals where cubic powers of $x$ are involved?
5. Can you use numerical methods to approximate the value of such integrals?

Tip: Trigonometric substitution is especially useful for integrals involving expressions like $a^2 - b^2x^2$, reducing them to simpler trigonometric forms.