The integral given is:

$\int \frac{x^3}{\sqrt{x^2 + 9}} \, dx$

The problem asks to use an appropriate trigonometric substitution to transform this into another form. Here's how we solve it step by step:

Step 1: Choose the trigonometric substitution

We have the term $\sqrt{x^2 + 9}$, so we can use the substitution:

$x = 3\tan\theta$

This substitution works because:

$x^2 + 9 = 9\tan^2\theta + 9 = 9(\tan^2\theta + 1) = 9\sec^2\theta$

Thus, $\sqrt{x^2 + 9} = 3\sec\theta$.

Step 2: Differentiate $x = 3\tan\theta$

Taking the derivative of both sides:

$dx = 3\sec^2\theta \, d\theta$

Step 3: Substitute into the integral

Now substitute $x = 3\tan\theta$ and $dx = 3\sec^2\theta \, d\theta$ into the integral:

$\int \frac{(3\tan\theta)^3}{3\sec\theta} \cdot 3\sec^2\theta \, d\theta$

This simplifies as follows:

$\int \frac{27\tan^3\theta}{\sec\theta} \cdot 3\sec^2\theta \, d\theta = 27 \int \tan^3\theta \sec\theta \, d\theta$

Step 4: Conclusion

Thus, the integral becomes:

$27 \int \tan^3\theta \sec^2\theta \, d\theta$

This matches option A in the choices.

Answer: The correct option is A.

Do you have any questions or would you like further details on any of the steps? Here are five related questions you might find helpful:

1. What are the common types of trigonometric substitutions used in integrals?
2. How do you simplify integrals involving powers of trigonometric functions like $\tan^3\theta$?
3. Why is the substitution $x = 3\tan\theta$ appropriate in this case?
4. What are other methods to solve integrals involving square roots and polynomials?
5. How does differentiation apply to trigonometric functions like $\sec^2\theta$ and $\tan\theta$?

Tip: When working with trigonometric substitutions, remember the identity $\tan^2\theta + 1 = \sec^2\theta$. This identity is crucial in simplifying the square root terms.