Evaluate `int(dx)/(cot^(2)x-1)`

Explanation

To evaluate the integral \( \int \frac{dx}{\cot^2 x - 1} \), we will follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting \( \cot^2 x \) in terms of sine and cosine:
\[ \cot^2 x = \frac{\cos^2 x}{\sin^2 x} \]
Thus, the integrand becomes:
\[ \frac{1}{\cot^2 x - 1} = \frac{1}{\frac{\cos^2 x}{\sin^2 x} - 1} = \frac{1}{\frac{\cos^2 x - \sin^2 x}{\sin^2 x}} = \frac{\sin^2 x}{\cos^2 x - \sin^2 x} \]

### Step 2: Use the identity for cosine
Using the identity \( \cos 2x = \cos^2 x - \sin^2 x \), we can rewrite the integral:
\[ \int \frac{\sin^2 x}{\cos^2 x - \sin^2 x} \, dx = \int \frac{\sin^2 x}{\cos 2x} \, dx \]

### Step 3: Substitute for sine squared
We can express \( \sin^2 x \) in terms of \( \cos 2x \):
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Substituting this into the integral gives:
\[ \int \frac{\frac{1 - \cos 2x}{2}}{\cos 2x} \, dx = \frac{1}{2} \int \frac{1 - \cos 2x}{\cos 2x} \, dx \]

### Step 4: Separate the integral
This can be separated into two integrals:
\[ \frac{1}{2} \left( \int \frac{1}{\cos 2x} \, dx - \int 1 \, dx \right) \]
This simplifies to:
\[ \frac{1}{2} \left( \int \sec 2x \, dx - x \right) \]

### Step 5: Evaluate the integral of secant
The integral of \( \sec 2x \) can be evaluated using the formula:
\[ \int \sec kx \, dx = \frac{1}{k} \ln |\sec kx + \tan kx| + C \]
For our case, \( k = 2 \):
\[ \int \sec 2x \, dx = \frac{1}{2} \ln |\sec 2x + \tan 2x| + C \]

### Step 6: Combine results
Putting everything together, we have:
\[ \frac{1}{2} \left( \frac{1}{2} \ln |\sec 2x + \tan 2x| - x \right) + C \]
This simplifies to:
\[ \frac{1}{4} \ln |\sec 2x + \tan 2x| - \frac{1}{2} x + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{dx}{\cot^2 x - 1} = \frac{1}{4} \ln |\sec 2x + \tan 2x| - \frac{1}{2} x + C \]