`int sec x tan x sqrt(tan^(2) x-4) dx`

Answer

To solve the integral \( \int \sec x \tan x \sqrt{\tan^2 x - 4} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \sec x \tan x \sqrt{\tan^2 x - 4} \, dx \]

### Step 2: Use Trigonometric Identity
Using the identity \( \tan^2 x = \sec^2 x - 1 \), we can rewrite \( \tan^2 x - 4 \) as:
\[ \tan^2 x - 4 = \sec^2 x - 1 - 4 = \sec^2 x - 5 \]
Thus, the integral becomes:
\[ \int \sec x \tan x \sqrt{\sec^2 x - 5} \, dx \]

### Step 3: Substitution
Let \( t = \sec x \). Then, the derivative \( dt = \sec x \tan x \, dx \). Therefore, we can rewrite the integral as:
\[ \int \sqrt{t^2 - 5} \, dt \]

### Step 4: Use a Standard Integral Formula
We recognize that \( \sqrt{t^2 - 5} \) can be integrated using the formula for the integral of the form \( \int \sqrt{x^2 - a^2} \, dx \):
\[ \int \sqrt{x^2 - a^2} \, dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log(x + \sqrt{x^2 - a^2}) + C \]
In our case, \( a^2 = 5 \).

### Step 5: Apply the Formula
Applying the formula, we have:
\[ \int \sqrt{t^2 - 5} \, dt = \frac{t}{2} \sqrt{t^2 - 5} - \frac{5}{2} \log(t + \sqrt{t^2 - 5}) + C \]

### Step 6: Substitute Back
Now we substitute back \( t = \sec x \):
\[ = \frac{\sec x}{2} \sqrt{\sec^2 x - 5} - \frac{5}{2} \log(\sec x + \sqrt{\sec^2 x - 5}) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \sec x \tan x \sqrt{\tan^2 x - 4} \, dx = \frac{\sec x}{2} \sqrt{\sec^2 x - 5} - \frac{5}{2} \log(\sec x + \sqrt{\sec^2 x - 5}) + C \]