`int (cos 2x)/(sin^(2) x) dx`

Answer

To solve the integral \( \int \frac{\cos 2x}{\sin^2 x} \, dx \), we can follow these steps:

### Step 1: Rewrite \( \cos 2x \)
We can use the double angle identity for cosine:
\[ \cos 2x = \cos^2 x - \sin^2 x \]
Thus, we can rewrite the integral as:
\[ \int \frac{\cos^2 x - \sin^2 x}{\sin^2 x} \, dx \]

### Step 2: Split the Integral
Now, we can separate the integral into two parts:
\[ \int \left( \frac{\cos^2 x}{\sin^2 x} - \frac{\sin^2 x}{\sin^2 x} \right) \, dx = \int \left( \cot^2 x - 1 \right) \, dx \]

### Step 3: Simplify the Integral
The expression simplifies to:
\[ \int \cot^2 x \, dx - \int 1 \, dx \]

### Step 4: Integrate Each Part
We know that:
\[ \int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx = -\cot x + x + C \]
Thus, we can combine the results:
\[ -\cot x + x - x + C = -\cot x + C \]

### Final Result
So, the final result of the integral is:
\[ -\cot x + C \]

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