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

Answer

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

### Step 1: Rewrite the integral using trigonometric identities
We know that \( \cos 2x = 1 - 2\sin^2 x \). Therefore, we can express \( \cos 2x \) in terms of \( \tan^2 x \):
\[ \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \]
Using this, we can rewrite the integral:
\[ I = \int \frac{1 + \cos 2x}{1 - \cos 2x} \, dx = \int \frac{1 + \frac{1 - \tan^2 x}{1 + \tan^2 x}}{1 - \frac{1 - \tan^2 x}{1 + \tan^2 x}} \, dx \]

### Step 2: Simplify the expression
Now, let's simplify the numerator and denominator:
- **Numerator**:
\[ 1 + \frac{1 - \tan^2 x}{1 + \tan^2 x} = \frac{(1 + \tan^2 x) + (1 - \tan^2 x)}{1 + \tan^2 x} = \frac{2}{1 + \tan^2 x} \]
- **Denominator**:
\[ 1 - \frac{1 - \tan^2 x}{1 + \tan^2 x} = \frac{(1 + \tan^2 x) - (1 - \tan^2 x)}{1 + \tan^2 x} = \frac{2\tan^2 x}{1 + \tan^2 x} \]

Thus, we can rewrite the integral as:
\[ I = \int \frac{\frac{2}{1 + \tan^2 x}}{\frac{2\tan^2 x}{1 + \tan^2 x}} \, dx = \int \frac{1}{\tan^2 x} \, dx \]

### Step 3: Rewrite in terms of cotangent
Since \( \frac{1}{\tan^2 x} = \cot^2 x \), we have:
\[ I = \int \cot^2 x \, dx \]

### Step 4: Integrate using the identity
We know that:
\[ \cot^2 x = \csc^2 x - 1 \]
Thus, we can split the integral:
\[ I = \int (\csc^2 x - 1) \, dx = \int \csc^2 x \, dx - \int 1 \, dx \]

### Step 5: Compute the integrals
The integral of \( \csc^2 x \) is:
\[ \int \csc^2 x \, dx = -\cot x \]
And the integral of \( 1 \) is:
\[ \int 1 \, dx = x \]

### Step 6: Combine the results
Putting it all together:
\[ I = -\cot x - x + C \]

### Final Answer
Thus, the final answer is:
\[ I = -\cot x - x + C \]