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

Answer

To solve the integral \( \int \frac{1 + \cos 2x}{1 - \cos 2x} \, dx \), we will use trigonometric identities and integration techniques. Here’s a step-by-step solution:

### Step 1: Simplify the integrand using trigonometric identities
We know the following trigonometric identities:
- \( 1 + \cos 2x = 2 \cos^2 x \)
- \( 1 - \cos 2x = 2 \sin^2 x \)

Using these identities, we can rewrite the integrand:
\[ \frac{1 + \cos 2x}{1 - \cos 2x} = \frac{2 \cos^2 x}{2 \sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x \]

### Step 2: Rewrite the integral
Now, we can rewrite the integral:
\[ \int \frac{1 + \cos 2x}{1 - \cos 2x} \, dx = \int \cot^2 x \, dx \]

### Step 3: Use the identity for cotangent
Recall that:
\[ \cot^2 x = \csc^2 x - 1 \]
Thus, we can express the integral as:
\[ \int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx \]

### Step 4: Integrate term by term
Now, we can integrate each term separately:
\[ \int \cot^2 x \, dx = \int \csc^2 x \, dx - \int 1 \, dx \]

The integrals are:
- \( \int \csc^2 x \, dx = -\cot x \)
- \( \int 1 \, dx = x \)

Putting it all together:
\[ \int \cot^2 x \, dx = -\cot x - x + C \]

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

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