To evaluate the integral

\[
I = \int \frac{\sec(x) \tan(x)}{\sec(x) - 1} \, dx
\]

we can proceed with the following steps:

### Step 1: Simplify the expression
Notice that the numerator is \(\sec(x) \tan(x)\), and we know that the derivative of \(\sec(x)\) is \(\sec(x) \tan(x)\). This suggests that substitution might work.

Let:
\[
u = \sec(x) - 1
\]
Then, the derivative of \(u\) is:
\[
du = \sec(x) \tan(x) \, dx
\]
which is exactly the numerator of the given integral.

### Step 2: Substitute and simplify
Using the substitution \(u = \sec(x) - 1\), the integral becomes:
\[
I = \int \frac{du}{u}
\]

### Step 3: Solve the new integral
This is a standard logarithmic integral, and its solution is:
\[
I = \ln |u| + C
\]
where \(C\) is the constant of integration.

### Step 4: Substitute back \(u = \sec(x) - 1\)
Finally, substitute back \(u = \sec(x) - 1\) to get:
\[
I = \ln |\sec(x) - 1| + C
\]

Thus, the solution to the integral is:
\[
\boxed{I = \ln |\sec(x) - 1| + C}
\]

Let me know if you would like further details!

### Relative Questions:
1. What is the derivative of \(\sec(x)\) and how does it help in this integration?
2. How does substitution simplify complex integrals like this one?
3. Can this method be applied to other trigonometric integrals?
4. What are the properties of the \(\sec(x)\) function that make it useful for integration?
5. How do you handle integrals involving \(\tan(x)\) and \(\sec(x)\) in other forms?

### Tip:
When dealing with integrals of trigonometric functions, always look for opportunities to use known derivatives, especially in the case of products involving \(\sec(x)\) and \(\tan(x)\).

\int \frac{\sec(x) \tan(x)}{\sec(x) - 1} \, dx

Solve the integral \int \frac{\sec(x) \tan(x)}{\sec(x) - 1} \, dx by using substitution. This step-by-step solution covers key trigonometric concepts, the derivative of sec(x), and logarithmic integration. Suitable for advanced high school students (Grades 11-12) or early college-level calculus.

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