We need to evaluate the integral:

$\int x \sqrt{6 - x^2} \, dx$

using substitution.

Solution:

1. Substitution: Let $u = 6 - x^2$. Then: $du = -2x \, dx \quad \text{or} \quad x \, dx = -\frac{1}{2} du.$

   The square root becomes: $\sqrt{6 - x^2} = \sqrt{u}.$

   Substituting into the integral: $\int x \sqrt{6 - x^2} \, dx = \int \sqrt{u} \cdot \left(-\frac{1}{2} \, du\right).$

2. Simplify: $\int x \sqrt{6 - x^2} \, dx = -\frac{1}{2} \int u^{1/2} \, du.$

3. Integrate: Using the power rule for integration: $\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}.$

   So: $-\frac{1}{2} \int u^{1/2} \, du = -\frac{1}{2} \cdot \frac{2}{3} u^{3/2} = -\frac{1}{3} u^{3/2}.$

4. Back-substitute $u = 6 - x^2$: $-\frac{1}{3} u^{3/2} = -\frac{1}{3} (6 - x^2)^{3/2}.$

5. Final answer: $\int x \sqrt{6 - x^2} \, dx = -\frac{1}{3} (6 - x^2)^{3/2} + C,$ where $C$ is the constant of integration.

Would you like a breakdown of any step, or have any questions?

Relative Questions:

1. How do you decide on the substitution $u = 6 - x^2$ for such integrals?
2. What is the power rule for integration, and how is it applied?
3. Can you explain why the $x \, dx$ term is replaced by $-\frac{1}{2} du$?
4. How does back-substitution work, and why is it necessary?
5. What are some other common substitution techniques for solving integrals?

Tip:

For integrals involving square roots of quadratic expressions, consider substitutions like $u = a - x^2$ or trigonometric substitutions for simplification.