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

Answer

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

### Step 1: Write the integral
We start with the integral:
\[ I = \int \frac{1}{1 + \cos 2x} \, dx \]

### Step 2: Use the half-angle identity
We know the half-angle identity for cosine:
\[ \cos 2x = 1 - 2\sin^2 x \]
Thus, we can rewrite \( 1 + \cos 2x \):
\[ 1 + \cos 2x = 1 + (1 - 2\sin^2 x) = 2 - 2\sin^2 x = 2(1 - \sin^2 x) = 2\cos^2 x \]

### Step 3: Substitute back into the integral
Now substituting this back into our integral, we have:
\[ I = \int \frac{1}{2 \cos^2 x} \, dx \]
This simplifies to:
\[ I = \frac{1}{2} \int \sec^2 x \, dx \]

### Step 4: Integrate \( \sec^2 x \)
The integral of \( \sec^2 x \) is a standard integral:
\[ \int \sec^2 x \, dx = \tan x + C \]
Thus, we can write:
\[ I = \frac{1}{2} (\tan x + C) \]

### Step 5: Write the final answer
Therefore, the final answer for the integral is:
\[ I = \frac{1}{2} \tan x + C \]