The given problem is an integral that needs to be solved using trigonometric substitution. Here's the expression:

$\int \frac{\sqrt{4x^2 - 25}}{x} \, dx$

The substitution provided is:

$x = \frac{5}{2} \sec(\theta)$

Step-by-Step Solution

1. Substitute $x = \frac{5}{2} \sec(\theta)$:

   First, differentiate $x = \frac{5}{2} \sec(\theta)$ to find $dx$:

   $dx = \frac{5}{2} \sec(\theta) \tan(\theta) \, d\theta$

2. Rewrite the integrand:

   Start by substituting $x = \frac{5}{2} \sec(\theta)$ into the expression inside the square root.

   $4x^2 - 25 = 4\left(\frac{5}{2} \sec(\theta)\right)^2 - 25$

   Simplifying the expression:

   $4x^2 - 25 = 4 \cdot \frac{25}{4} \sec^2(\theta) - 25 = 25 \sec^2(\theta) - 25 = 25(\sec^2(\theta) - 1)$

   Using the trigonometric identity $\sec^2(\theta) - 1 = \tan^2(\theta)$, we get:

   $4x^2 - 25 = 25 \tan^2(\theta)$

   Therefore, the square root becomes:

   $\sqrt{4x^2 - 25} = \sqrt{25 \tan^2(\theta)} = 5 \tan(\theta)$

3. Substitute into the integral:

   Now substitute both $\sqrt{4x^2 - 25}$ and $dx$ into the integral:

   $\int \frac{\sqrt{4x^2 - 25}}{x} \, dx = \int \frac{5 \tan(\theta)}{\frac{5}{2} \sec(\theta)} \cdot \frac{5}{2} \sec(\theta) \tan(\theta) \, d\theta$

   Simplify the expression:

   $= \int 5 \tan^2(\theta) \, d\theta$

4. Solve the integral:

   We can use the identity $\tan^2(\theta) = \sec^2(\theta) - 1$ to simplify the integral:

   $\int 5 \tan^2(\theta) \, d\theta = 5 \int (\sec^2(\theta) - 1) \, d\theta$

   This separates into two integrals:

   $5 \int \sec^2(\theta) \, d\theta - 5 \int 1 \, d\theta$

   These are straightforward to solve:

   $5 \left(\tan(\theta)\right) - 5\theta + C$

   So the solution in terms of $\theta$ is:

   $5 \tan(\theta) - 5\theta + C$

5. Substitute back $x = \frac{5}{2} \sec(\theta)$:

   Recall that $x = \frac{5}{2} \sec(\theta)$, so $\sec(\theta) = \frac{2x}{5}$. Therefore:

   $\theta = \sec^{-1}\left(\frac{2x}{5}\right)$

   Also, $\tan(\theta)$ can be found from the right triangle relationship:

   $\tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\left(\frac{2x}{5}\right)^2 - 1} = \frac{\sqrt{4x^2 - 25}}{5}$

   Thus, the final solution becomes:

   $\frac{\sqrt{4x^2 - 25}}{x} - 5 \sec^{-1}\left(\frac{2x}{5}\right) + C$

This is the evaluated integral using trigonometric substitution.

Do you want more details or have any questions?

Here are some related questions:

1. What are the steps to solve other integrals using trigonometric substitution?
2. How does the identity $\sec^2(\theta) - 1 = \tan^2(\theta)$ help simplify integrals?
3. What are other common trigonometric substitutions used for integrals?
4. How can you derive the inverse trigonometric functions for specific substitutions?
5. What is the difference between solving integrals with trigonometric substitution versus other methods?

Tip: Trigonometric identities like $\sec^2(\theta) - 1 = \tan^2(\theta)$ are essential when simplifying expressions during integration steps.