`int(cos2x-1)/(cos2x+1)dx` is equal to

Options

A. tanx-x+c
B. x+tanx+c
C. x-tanx+c
D. `-x-cotx+c`

Explanation

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

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ \int \frac{\cos(2x) - 1}{\cos(2x) + 1} \, dx = \int \frac{-(1 - \cos(2x))}{\cos(2x) + 1} \, dx \]
This allows us to factor out a negative sign:
\[ = -\int \frac{1 - \cos(2x)}{\cos(2x) + 1} \, dx \]

### Step 2: Use Trigonometric Identities
Using the identity \(1 - \cos(2x) = 2\sin^2(x)\) and \(\cos(2x) = 2\cos^2(x) - 1\), we can rewrite the integral:
\[ = -\int \frac{2\sin^2(x)}{2\cos^2(x)} \, dx \]
This simplifies to:
\[ = -\int \frac{\sin^2(x)}{\cos^2(x)} \, dx = -\int \tan^2(x) \, dx \]

### Step 3: Integrate \(\tan^2(x)\)
We know that \(\tan^2(x) = \sec^2(x) - 1\), so we can rewrite the integral:
\[ = -\int (\sec^2(x) - 1) \, dx \]
This separates into two integrals:
\[ = -\left(\int \sec^2(x) \, dx - \int 1 \, dx\right) \]

### Step 4: Compute the Integrals
The integral of \(\sec^2(x)\) is \(\tan(x)\), and the integral of \(1\) is \(x\):
\[ = -\left(\tan(x) - x\right) + C \]
Thus, we have:
\[ = x - \tan(x) + C \]

### Final Result
The final result of the integral is:
\[ \int \frac{\cos(2x) - 1}{\cos(2x) + 1} \, dx = x - \tan(x) + C \]