To rewrite the given integral using trigonometric substitution, we typically use a substitution of the form:

$x = 7 \sin(\theta)$

This substitution is appropriate because it simplifies the square root term in the denominator.

Let's go through the steps of the substitution process for the integral:

Given: $\int \frac{x^2}{\sqrt{49 - x^2}} \, dx$

Step 1: Trigonometric Substitution

Let: $x = 7 \sin(\theta)$ Then: $dx = 7 \cos(\theta) \, d\theta$ And: $\sqrt{49 - x^2} = \sqrt{49 - 49 \sin^2(\theta)} = \sqrt{49(1 - \sin^2(\theta))} = \sqrt{49 \cos^2(\theta)} = 7 \cos(\theta)$

Step 2: Substitute into the Integral

Now, substitute these expressions into the original integral:

$\int \frac{x^2}{\sqrt{49 - x^2}} \, dx = \int \frac{(7 \sin(\theta))^2}{7 \cos(\theta)} \cdot 7 \cos(\theta) \, d\theta$ Simplifying:

$= \int \frac{49 \sin^2(\theta)}{7 \cos(\theta)} \cdot 7 \cos(\theta) \, d\theta$ $= \int 49 \sin^2(\theta) \, d\theta$

Step 3: Final Expression

So, the integral becomes: $\int 49 \sin^2(\theta) \, d\theta$

This is the rewritten integral using the substitution $x = 7 \sin(\theta)$.

Would you like to proceed with the solution for this integral, or do you have any other questions on this topic?

Here are some related questions you might find useful:

1. How do you integrate $\sin^2(\theta)$?
2. What are other common trigonometric substitutions used in integrals?
3. How do you perform inverse trigonometric substitutions?
4. What is the purpose of using trigonometric substitution in integrals?
5. Can you solve definite integrals using the substitution $x = 7 \sin(\theta)$?

Tip: When performing trigonometric substitution, always pay attention to the range of $\theta$ to ensure the correct limits when solving definite integrals.