`int (2x-sin 2x)/(1-cos 2x) dx`

Answer

To solve the integral \( \int \frac{2x - \sin 2x}{1 - \cos 2x} \, dx \), we can follow these steps:

### Step 1: Simplify the Integral
We start with the integral:
\[ \int \frac{2x - \sin 2x}{1 - \cos 2x} \, dx \]

Using the trigonometric identities:
- \( \sin 2x = 2 \sin x \cos x \)
- \( 1 - \cos 2x = 2 \sin^2 x \)

We can rewrite the integral as:
\[ \int \frac{2x - 2 \sin x \cos x}{2 \sin^2 x} \, dx \]
This simplifies to:
\[ \int \frac{x - \sin x \cos x}{\sin^2 x} \, dx \]

### Step 2: Split the Integral
Now, we can split the integral into two parts:
\[ \int \frac{x}{\sin^2 x} \, dx - \int \frac{\sin x \cos x}{\sin^2 x} \, dx \]
This further simplifies to:
\[ \int x \csc^2 x \, dx - \int \cot x \, dx \]

### Step 3: Integrate Each Part
1. **Integrate \( \int x \csc^2 x \, dx \)** using integration by parts:
- Let \( u = x \) and \( dv = \csc^2 x \, dx \)
- Then \( du = dx \) and \( v = -\cot x \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int x \csc^2 x \, dx = -x \cot x - \int -\cot x \, dx \]
This simplifies to:
\[ -x \cot x + \int \cot x \, dx \]

2. **Integrate \( \int \cot x \, dx \)**:
\[ \int \cot x \, dx = \ln |\sin x| + C \]

### Step 4: Combine the Results
Now, substituting back into our integral:
\[ \int \frac{2x - \sin 2x}{1 - \cos 2x} \, dx = -x \cot x + \int \cot x \, dx - \int \cot x \, dx \]
The \( \int \cot x \, dx \) terms cancel out, leaving us with:
\[ -x \cot x + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{2x - \sin 2x}{1 - \cos 2x} \, dx = -x \cot x + C \]