`int ((1-cos2x))/((1+cos 2x)) dx = ?`

Options

A. `tanx + x + C`
B. `tanx - x + C`
C. `-tan x + x + C`
D. `- tanx- x + C`

Answer

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

### Step 1: Rewrite the integrand
We know that \(\cos 2x = 2\cos^2 x - 1\). Thus, we can express \(1 + \cos 2x\) as:
\[ 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x \]
And \(1 - \cos 2x\) can be expressed as:
\[ 1 - \cos 2x = 1 - (2\cos^2 x - 1) = 2 - 2\cos^2 x = 2(1 - \cos^2 x) = 2\sin^2 x \]

### Step 2: Substitute into the integral
Now substituting these into the integral, we have:
\[ \int \frac{1 - \cos 2x}{1 + \cos 2x} \, dx = \int \frac{2\sin^2 x}{2\cos^2 x} \, dx = \int \frac{\sin^2 x}{\cos^2 x} \, dx = \int \tan^2 x \, dx \]

### Step 3: Use the identity for \(\tan^2 x\)
We know that:
\[ \tan^2 x = \sec^2 x - 1 \]
Thus, we can rewrite the integral as:
\[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx \]

### Step 4: Integrate
Now we can integrate:
\[ \int \tan^2 x \, dx = \int \sec^2 x \, dx - \int 1 \, dx = \tan x - x + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{1 - \cos 2x}{1 + \cos 2x} \, dx = \tan x - x + C \]