To solve the integral

$\int \frac{x^2}{\sqrt{1 - x^2}} \, dx,$

we can use the following method. Let's perform the substitution $x = \sin(\theta)$. Then, $dx = \cos(\theta) \, d\theta$ and $\sqrt{1 - x^2} = \sqrt{1 - \sin^2(\theta)} = \cos(\theta)$.

Substitute these into the integral:

$\int \frac{x^2}{\sqrt{1 - x^2}} \, dx = \int \frac{\sin^2(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^2(\theta) \, d\theta.$

Now, use the trigonometric identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$:

$\int \sin^2(\theta) \, d\theta = \int \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int (1 - \cos(2\theta)) \, d\theta.$

This integral can be split into two simpler integrals:

$\frac{1}{2} \left( \int 1 \, d\theta - \int \cos(2\theta) \, d\theta \right).$

Integrate each term:

$\int 1 \, d\theta = \theta \quad \text{and} \quad \int \cos(2\theta) \, d\theta = \frac{1}{2} \sin(2\theta).$

Thus, we have:

$\frac{1}{2} \left( \theta - \frac{1}{2} \sin(2\theta) \right) + C.$

Using the double-angle identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, we get:

$\frac{1}{2} \left( \theta - \frac{1}{2} \cdot 2 \sin(\theta) \cos(\theta) \right) + C = \frac{1}{2} \left( \theta - \sin(\theta) \cos(\theta) \right) + C.$

Recall that $x = \sin(\theta)$, so $\theta = \arcsin(x)$, and $\cos(\theta) = \sqrt{1 - x^2}$. Substituting these back in, we get:

$\frac{1}{2} \left( \arcsin(x) - x \sqrt{1 - x^2} \right) + C.$

Therefore, the solution to the integral is:

$\int \frac{x^2}{\sqrt{1 - x^2}} \, dx = \frac{1}{2} \left( \arcsin(x) - x \sqrt{1 - x^2} \right) + C.$

Would you like more details or have any questions about this solution?

Here are 8 related questions you might find interesting:

1. How do you handle integrals involving trigonometric functions?
2. What are common substitutions used in integration?
3. How can trigonometric identities simplify integration?
4. What is the process of reversing trigonometric substitutions?
5. How do you integrate functions involving square roots?
6. Can you solve $\int \frac{x}{\sqrt{1 - x^2}} \, dx$?
7. How do you deal with definite integrals using trigonometric substitution?
8. What are the applications of integrals involving trigonometric functions?

Tip: Always consider substitution and trigonometric identities when faced with complex integrals involving polynomials and roots.