`int ((1-cotx))/((1+cotx))dx.`

Answer

To solve the integral \( I = \int \frac{1 - \cot x}{1 + \cot x} \, dx \), we will follow these steps:

### Step 1: Rewrite cotangent in terms of sine and cosine
We know that \( \cot x = \frac{\cos x}{\sin x} \). Therefore, we can rewrite the integral as:
\[ I = \int \frac{1 - \frac{\cos x}{\sin x}}{1 + \frac{\cos x}{\sin x}} \, dx \]

### Step 2: Simplify the expression
To simplify the fraction, we will multiply the numerator and the denominator by \( \sin x \):
\[ I = \int \frac{\sin x - \cos x}{\sin x + \cos x} \, dx \]

### Step 3: Use substitution
Let \( t = \sin x + \cos x \). Then, we need to find \( dt \):
\[ dt = \cos x \, dx - \sin x \, dx = (\cos x - \sin x) \, dx \]
This implies:
\[ dx = \frac{dt}{\cos x - \sin x} \]

### Step 4: Express the numerator in terms of \( t \)
From our substitution \( t = \sin x + \cos x \), we can express \( \sin x - \cos x \) in terms of \( t \):
\[ \sin x - \cos x = -(\cos x - \sin x) \]
Thus, we can rewrite the integral:
\[ I = \int \frac{-(\cos x - \sin x)}{t} \cdot \frac{dt}{\cos x - \sin x} \]
This simplifies to:
\[ I = -\int \frac{1}{t} \, dt \]

### Step 5: Integrate
The integral of \( \frac{1}{t} \) is:
\[ I = -\ln |t| + C \]

### Step 6: Substitute back for \( t \)
Now we substitute back \( t = \sin x + \cos x \):
\[ I = -\ln |\sin x + \cos x| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = -\ln |\sin x + \cos x| + C \]

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