` int (1)/(1-cot x)`

Options

A. `1/2log|sinx-cosx|+1/2x+C`
B. `1/2log|sinx-cosx|+1/2x+C`
C. `1/2log|sinx+cosx|-1/2x+C`
D. `1/2log|sinx-cosx|-1/2x+C`

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting \( \cot x \) in terms of sine and cosine:
\[ \cot x = \frac{\cos x}{\sin x} \]
Thus, we can express the integral as:
\[ I = \int \frac{1}{1 - \frac{\cos x}{\sin x}} \, dx \]

### Step 2: Simplify the integrand
Next, we simplify the expression in the integrand:
\[ I = \int \frac{1}{\frac{\sin x - \cos x}{\sin x}} \, dx = \int \frac{\sin x}{\sin x - \cos x} \, dx \]

### Step 3: Split the integral
Now, we can multiply and divide the integrand by 2:
\[ I = \frac{1}{2} \int \frac{2 \sin x}{\sin x - \cos x} \, dx \]
We can express \( 2 \sin x \) as \( \sin x + \sin x \):
\[ I = \frac{1}{2} \int \left( \frac{\sin x - \cos x}{\sin x - \cos x} + \frac{\sin x + \cos x}{\sin x - \cos x} \right) \, dx \]

### Step 4: Separate the integrals
This allows us to separate the integral:
\[ I = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx \]

### Step 5: Evaluate the first integral
The first integral is straightforward:
\[ \frac{1}{2} \int 1 \, dx = \frac{1}{2} x \]

### Step 6: Substitute for the second integral
For the second integral, we will use the substitution:
Let \( t = \sin x - \cos x \). Then, differentiating gives:
\[ dt = (\cos x + \sin x) \, dx \]
Thus, we can rewrite the second integral:
\[ \int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx = \int \frac{1}{t} \, dt \]

### Step 7: Evaluate the second integral
The integral of \( \frac{1}{t} \) is:
\[ \int \frac{1}{t} \, dt = \log |t| + C = \log |\sin x - \cos x| + C \]

### Step 8: Combine the results
Combining both parts, we have:
\[ I = \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C \]

### Final Result
Thus, the final answer is:
\[ I = \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C \]