We are given the following integral and substitution:

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

with the substitution $x = 3 \sec(\theta)$.

Step-by-Step Solution:

1. Substitute $x = 3 \sec(\theta)$:

   $dx = 3 \sec(\theta) \tan(\theta) \, d\theta$ Also, we need to compute $\sqrt{x^2 - 9}$:

   $x^2 = 9 \sec^2(\theta)$ $x^2 - 9 = 9 (\sec^2(\theta) - 1) = 9 \tan^2(\theta)$ Therefore:

   $\sqrt{x^2 - 9} = 3 \tan(\theta)$

2. Substitute into the integral:

   Now, the integral becomes:

   $\int \frac{3 \tan(\theta)}{3 \sec(\theta)} \cdot 3 \sec(\theta) \tan(\theta) \, d\theta$ Simplify the expression:

   $\int \tan^2(\theta) \cdot 3 \, d\theta$

3. Simplify and integrate:

   Recall the identity:

   $\tan^2(\theta) = \sec^2(\theta) - 1$ So, the integral becomes:

   $3 \int (\sec^2(\theta) - 1) \, d\theta$ This simplifies to:

   $3 \left( \int \sec^2(\theta) \, d\theta - \int 1 \, d\theta \right)$

   The integrals of $\sec^2(\theta)$ and 1 are:

   $\int \sec^2(\theta) \, d\theta = \tan(\theta), \quad \int 1 \, d\theta = \theta$

   Therefore, the integral becomes:

   $3 (\tan(\theta) - \theta) + C$

4. Return to $x$-terms:

   From the substitution $x = 3 \sec(\theta)$, we know that:

   $\sec(\theta) = \frac{x}{3}, \quad \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\frac{x^2}{9} - 1} = \frac{\sqrt{x^2 - 9}}{3}$

   Also:

   $\theta = \sec^{-1}\left(\frac{x}{3}\right)$

   Thus, the solution in terms of $x$ is:

   $\int \frac{\sqrt{x^2 - 9}}{x} \, dx = \sqrt{x^2 - 9} - 3 \sec^{-1}\left(\frac{x}{3}\right) + C$

This is the final result.

Do you have any questions or would you like further details on any steps?

Related Questions:

1. How does trigonometric substitution help in evaluating integrals?
2. What is the general strategy for solving integrals involving square roots of quadratic expressions?
3. Can you explain the derivation of the identity $\sec^2(\theta) - 1 = \tan^2(\theta)$?
4. How do you handle integrals involving $\sec(\theta)$ and $\tan(\theta)$?
5. How would the solution change if the integral involved $x^2 + 9$ instead of $x^2 - 9$?

Tip:

When dealing with integrals that involve square roots of quadratic expressions, trigonometric substitutions are often the most effective approach to simplify the problem.