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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:
   \[
   = -3 \cdot \left( \frac{\pi}{4} - \frac{1}{2} \right)
   = -\frac{3\pi}{4} + \frac{3}{2}
   \]

### Final Answer:
\[
-\frac{3\pi}{4} + \frac{3}{2}
\]

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

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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.

Evaluate the integral: \(\int_{0}^{\frac{\sqrt{2}}{2}} \frac{-6x^2}{\sqrt{1 - x^2}} \, dx\)

This problem involves evaluating an integral using trigonometric substitution. It focuses on concepts from Integral Calculus and applies the Pythagorean identity and the Fundamental Theorem of Calculus. A detailed, step-by-step solution is provided, suitable for college-level students.

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