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

Options

A. `log(sinx)-log(1-cotx)+c`
B. `log(sinx)+log(1-cotx)+c`
C. `log(cotx-1)+c`
D. `log(1+cotx)+c`

Answer

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

### Step 1: Rewrite the integrand
We start with the integral:
\[ \int \frac{1 + \cot x}{1 - \cot x} \, dx \]
Recall that \(\cot x = \frac{\cos x}{\sin x}\). Thus, we can rewrite the integrand:
\[ \frac{1 + \cot x}{1 - \cot x} = \frac{1 + \frac{\cos x}{\sin x}}{1 - \frac{\cos x}{\sin x}} = \frac{\sin x + \cos x}{\sin x - \cos x} \]

### Step 2: Substitute and simplify
Now, we can rewrite the integral:
\[ \int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx \]

### Step 3: Use substitution
Let \(u = \sin x - \cos x\). Then, we differentiate \(u\):
\[ du = (\cos x + \sin x) \, dx \]
This means:
\[ dx = \frac{du}{\cos x + \sin x} \]

### Step 4: Substitute back into the integral
Now we can substitute \(u\) and \(dx\) into the integral:
\[ \int \frac{\sin x + \cos x}{u} \cdot \frac{du}{\cos x + \sin x} = \int \frac{1}{u} \, du \]

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

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