To evaluate the integral

$\int \frac{x^2}{\sqrt{9 - x^2}} \, dx,$

let's follow these steps:

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Step 1: Substitute for the square root term

We use the trigonometric substitution:

$x = 3 \sin \theta \quad \text{so that} \quad dx = 3 \cos \theta \, d\theta \quad \text{and} \quad \sqrt{9 - x^2} = 3 \cos \theta.$

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Step 2: Substitute in the integral

Replacing $x^2$, $dx$, and $\sqrt{9 - x^2}$ using the substitution:

$x^2 = 9 \sin^2 \theta, \quad dx = 3 \cos \theta \, d\theta, \quad \sqrt{9 - x^2} = 3 \cos \theta,$

the integral becomes:

$\int \frac{9 \sin^2 \theta}{3 \cos \theta} \cdot 3 \cos \theta \, d\theta.$

Simplify the terms:

$\int 9 \sin^2 \theta \, d\theta.$

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Step 3: Simplify using trigonometric identities

Recall the identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$. Substitute this:

$\int 9 \cdot \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{9}{2} \int (1 - \cos(2\theta)) \, d\theta.$

Distribute the integral:

$\frac{9}{2} \int 1 \, d\theta - \frac{9}{2} \int \cos(2\theta) \, d\theta.$

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Step 4: Integrate each term

1. For $\int 1 \, d\theta$, the result is $\theta$.
2. For $\int \cos(2\theta) \, d\theta$, use the substitution $u = 2\theta$, giving the result $\frac{\sin(2\theta)}{2}$.

Thus:

$\frac{9}{2} \int (1 - \cos(2\theta)) \, d\theta = \frac{9}{2} \left(\theta - \frac{\sin(2\theta)}{2}\right).$

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Step 5: Back-substitute $\theta$ in terms of $x$

Recall that $x = 3 \sin \theta$, so $\sin \theta = \frac{x}{3}$ and $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{x^2}{9}} = \frac{\sqrt{9 - x^2}}{3}$.

Also, $\sin(2\theta) = 2 \sin \theta \cos \theta = 2 \cdot \frac{x}{3} \cdot \frac{\sqrt{9 - x^2}}{3} = \frac{2x \sqrt{9 - x^2}}{9}$.

Substituting back:

$\theta = \arcsin\left(\frac{x}{3}\right), \quad \sin(2\theta) = \frac{2x \sqrt{9 - x^2}}{9}.$

The integral becomes:

$\frac{9}{2} \left(\arcsin\left(\frac{x}{3}\right) - \frac{1}{2} \cdot \frac{2x \sqrt{9 - x^2}}{9}\right) = \frac{9}{2} \arcsin\left(\frac{x}{3}\right) - x \sqrt{9 - x^2}.$

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Final Answer:

$\int \frac{x^2}{\sqrt{9 - x^2}} \, dx = \frac{9}{2} \arcsin\left(\frac{x}{3}\right) - x \sqrt{9 - x^2} + C,$ where $C$ is the constant of integration.

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Would you like a step-by-step breakdown of any part?

Related Questions:

1. What is the geometric meaning of $\arcsin\left(\frac{x}{3}\right)$?
2. How is the trigonometric substitution $x = 3 \sin \theta$ chosen for this integral?
3. What alternative methods could solve this integral?
4. Why is $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$ used in simplification?
5. What does the integral represent geometrically or physically?

Tip:

When performing trigonometric substitution, always verify the range of the variable to ensure the substitution aligns with the function's domain.