`int tan^(-1) ((sin2x)/(1+cos 2x))dx`

Answer

To solve the integral \( \int \tan^{-1} \left( \frac{\sin 2x}{1 + \cos 2x} \right) dx \), we will use trigonometric identities to simplify the expression inside the integral.

### Step-by-Step Solution:

1. **Use Trigonometric Identities**:
We know the following identities:
- \( \sin 2x = 2 \sin x \cos x \)
- \( \cos 2x = 2 \cos^2 x - 1 \)

From the cosine identity, we can express \( 1 + \cos 2x \):
\[ 1 + \cos 2x = 1 + (2 \cos^2 x - 1) = 2 \cos^2 x \]

2. **Substitute into the Integral**:
Substitute \( \sin 2x \) and \( 1 + \cos 2x \) into the integral:
\[ \int \tan^{-1} \left( \frac{2 \sin x \cos x}{2 \cos^2 x} \right) dx \]

3. **Simplify the Expression**:
The expression simplifies as follows:
\[ \frac{2 \sin x \cos x}{2 \cos^2 x} = \frac{\sin x}{\cos x} = \tan x \]
Therefore, the integral becomes:
\[ \int \tan^{-1}(\tan x) \, dx \]

4. **Evaluate the Integral**:
Since \( \tan^{-1}(\tan x) = x \) for \( x \) in the principal range of \( \tan^{-1} \), we have:
\[ \int x \, dx \]

5. **Integrate**:
The integral of \( x \) is:
\[ \frac{x^2}{2} + C \]

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