The given integral is:

$\int_{0}^{\frac{\sqrt{2}}{2}} \frac{-6x^2}{\sqrt{1 - x^2}} \, dx$

Approach:

1. Substitution: A common method for integrals involving $\sqrt{1 - x^2}$ is trigonometric substitution, typically $x = \sin(\theta)$, because $\sin^2(\theta) + \cos^2(\theta) = 1$.

2. Substitution Details: Let $x = \sin(\theta)$. Then: $dx = \cos(\theta) d\theta$ and $\sqrt{1 - x^2} = \sqrt{1 - \sin^2(\theta)} = \cos(\theta)$

   The limits of integration also change accordingly. When $x = 0$, $\theta = 0$, and when $x = \frac{\sqrt{2}}{2}$, $\theta = \frac{\pi}{4}$.

3. Transforming the Integral: Using the substitution $x = \sin(\theta)$, the integral becomes: $\int_{0}^{\frac{\pi}{4}} \frac{-6\sin^2(\theta)}{\cos(\theta)} \cdot \cos(\theta) \, d\theta$

   Simplifying: $\int_{0}^{\frac{\pi}{4}} -6\sin^2(\theta) \, d\theta$

4. Solving the Integral: To integrate $\sin^2(\theta)$, we use the identity: $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$ Therefore, the integral becomes: $\int_{0}^{\frac{\pi}{4}} -6 \cdot \frac{1 - \cos(2\theta)}{2} \, d\theta$ $= -3 \int_{0}^{\frac{\pi}{4}} (1 - \cos(2\theta)) \, d\theta$ $= -3 \left[ \theta - \frac{\sin(2\theta)}{2} \right]_{0}^{\frac{\pi}{4}}$

5. Evaluating the Boundaries: At $\theta = \frac{\pi}{4}$: $\frac{\sin(2\cdot\frac{\pi}{4})}{2} = \frac{\sin(\frac{\pi}{2})}{2} = \frac{1}{2}$ So: $\left( \frac{\pi}{4} - \frac{1}{2} \right)$

   At $\theta = 0$, both terms are zero.

   Therefore, the integral evaluates to: $-3 \left( \frac{\pi}{4} - \frac{1}{2} \right)$ Simplifying:

   = -\frac{3\pi}{4} + \frac{3}{2}$$

Final Answer:

$-\frac{3\pi}{4} + \frac{3}{2}$

Would you like further explanation or details on any part of the solution?

Here are 5 related questions:

1. How do you apply trigonometric substitution for integrals involving $\sqrt{1 - x^2}$?
2. Can you solve an integral with $\cos^2(\theta)$ using a similar approach?
3. How does the Pythagorean identity simplify integral calculations?
4. How does changing the limits of integration work in substitution methods?
5. What is the integral of $\sin(2\theta)$, and how is it derived?

Tip: For integrals involving square roots of quadratic terms like $\sqrt{1 - x^2}$, trigonometric substitution is often a powerful method to simplify the integral.