`int(1-cosx)/(cosx(1+cosx))dx`

Explanation

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

### Step 1: Rewrite the integrand
We start with the expression:
\[ \frac{1 - \cos x}{\cos x (1 + \cos x)} \]
We can rewrite \(1 - \cos x\) as:
\[ 1 - \cos x = (1 + \cos x) - 2\cos x \]
This allows us to express the integrand as:
\[ \frac{(1 + \cos x) - 2\cos x}{\cos x (1 + \cos x)} = \frac{1 + \cos x}{\cos x (1 + \cos x)} - \frac{2\cos x}{\cos x (1 + \cos x)} \]
This simplifies to:
\[ \frac{1}{\cos x} - \frac{2}{1 + \cos x} \]

### Step 2: Separate the integral
Now we can separate the integral:
\[ \int \left( \frac{1}{\cos x} - \frac{2}{1 + \cos x} \right) \, dx = \int \sec x \, dx - 2 \int \frac{1}{1 + \cos x} \, dx \]

### Step 3: Solve the first integral
The integral of \(\sec x\) is:
\[ \int \sec x \, dx = \ln | \sec x + \tan x | + C_1 \]

### Step 4: Solve the second integral
For the integral \(\int \frac{1}{1 + \cos x} \, dx\), we can use the identity \(1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)\):
\[ \int \frac{1}{1 + \cos x} \, dx = \int \frac{1}{2 \cos^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \sec^2\left(\frac{x}{2}\right) \, dx \]
The integral of \(\sec^2\) is:
\[ \int \sec^2\left(\frac{x}{2}\right) \, dx = 2 \tan\left(\frac{x}{2}\right) + C_2 \]
Thus, we have:
\[ \int \frac{1}{1 + \cos x} \, dx = \tan\left(\frac{x}{2}\right) + C_2 \]

### Step 5: Combine the results
Putting it all together, we have:
\[ \int \frac{1 - \cos x}{\cos x (1 + \cos x)} \, dx = \ln | \sec x + \tan x | - 2 \tan\left(\frac{x}{2}\right) + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{1 - \cos x}{\cos x (1 + \cos x)} \, dx = \ln | \sec x + \tan x | - 2 \tan\left(\frac{x}{2}\right) + C \]