1. **State the problem:** We need to evaluate the definite integral $$\int_0^{\frac{\sqrt{2}}{2}} \frac{-3x^2}{\sqrt{1 - x^2}} \, dx$$ 2. **Recall the formula and substitution:** The integral involves a function with a square root in the denominator. A common substitution for expressions involving $\sqrt{1 - x^2}$ is to use a trigonometric substitution or recognize the derivative of $\sqrt{1 - x^2}$. 3. **Rewrite the integral:** $$\int_0^{\frac{\sqrt{2}}{2}} \frac{-3x^2}{\sqrt{1 - x^2}} \, dx = -3 \int_0^{\frac{\sqrt{2}}{2}} \frac{x^2}{\sqrt{1 - x^2}} \, dx$$ 4. **Use substitution:** Let $$u = 1 - x^2 \implies du = -2x \, dx \implies -\frac{1}{2} du = x \, dx$$ We need to express $x^2 \, dx$ in terms of $u$ and $du$. Notice that $x^2 = 1 - u$. Rewrite the integral: $$-3 \int_0^{\frac{\sqrt{2}}{2}} \frac{x^2}{\sqrt{1 - x^2}} \, dx = -3 \int \frac{1 - u}{\sqrt{u}} \cdot \frac{-1}{2x} du$$ But this is complicated because of the $x$ in the denominator. Instead, use integration by parts or rewrite the integral differently. 5. **Alternative approach:** Write $x^2 = (1 - (1 - x^2)) = 1 - u$ and rewrite the integral as $$-3 \int_0^{\frac{\sqrt{2}}{2}} \frac{x^2}{\sqrt{1 - x^2}} \, dx = -3 \int_0^{\frac{\sqrt{2}}{2}} \frac{1 - u}{\sqrt{u}} \, dx$$ But $dx$ is still in terms of $x$, so better to use a trigonometric substitution. 6. **Trigonometric substitution:** Let $$x = \sin \theta \implies dx = \cos \theta \, d\theta$$ When $x=0$, $\theta=0$; when $x=\frac{\sqrt{2}}{2}$, $\theta=\frac{\pi}{4}$. Rewrite the integral: $$-3 \int_0^{\frac{\sqrt{2}}{2}} \frac{x^2}{\sqrt{1 - x^2}} \, dx = -3 \int_0^{\frac{\pi}{4}} \frac{\sin^2 \theta}{\sqrt{1 - \sin^2 \theta}} \cos \theta \, d\theta$$ Since $\sqrt{1 - \sin^2 \theta} = \cos \theta$, the integral becomes $$-3 \int_0^{\frac{\pi}{4}} \frac{\sin^2 \theta}{\cos \theta} \cos \theta \, d\theta = -3 \int_0^{\frac{\pi}{4}} \sin^2 \theta \, d\theta$$ 7. **Simplify the integral:** $$-3 \int_0^{\frac{\pi}{4}} \sin^2 \theta \, d\theta$$ Use the identity: $$\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$$ So the integral becomes $$-3 \int_0^{\frac{\pi}{4}} \frac{1 - \cos(2\theta)}{2} \, d\theta = -\frac{3}{2} \int_0^{\frac{\pi}{4}} (1 - \cos(2\theta)) \, d\theta$$ 8. **Integrate term by term:** $$-\frac{3}{2} \left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{4}} = -\frac{3}{2} \left( \frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2} - 0 \right)$$ Since $\sin(\frac{\pi}{2}) = 1$, this is $$-\frac{3}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right) = -\frac{3}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right)$$ 9. **Final simplification:** $$-\frac{3}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right) = -\frac{3}{2} \cdot \frac{\pi}{4} + \frac{3}{2} \cdot \frac{1}{2} = -\frac{3\pi}{8} + \frac{3}{4}$$ **Answer:** $$\boxed{\frac{3}{4} - \frac{3\pi}{8}}$$