`int (6dx)/(sin^2x(1-cot^2x)`= (where c is arbitrary constant)

Options

A. `1/3 ln |(1 -cot x)/(1 + cot x)|+c`
B. `6 log_e |(1 -cot x)/(1 + cot x)|+c`
C. `1/3 ln |(1 + cot x)/(1 - cot x)|+c`
D. `1/3 log_e |(cot x - 1)/(cot x + 1)|+c`

Answer

To solve the integral

\[ \int \frac{6 \, dx}{\sin^2 x (1 - \cot^2 x)} \]

we can follow these steps:

### Step 1: Simplify the integrand
We know that

\[ 1 - \cot^2 x = 1 - \frac{\cos^2 x}{\sin^2 x} = \frac{\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{\sin^2 x - \cos^2 x}{\sin^2 x} \]

Thus, we can rewrite the integral as:

\[ \int \frac{6 \, dx}{\sin^2 x \cdot \frac{\sin^2 x - \cos^2 x}{\sin^2 x}} = \int \frac{6 \sin^2 x \, dx}{\sin^2 x - \cos^2 x} \]

### Step 2: Use a substitution
Let \( t = \cot x \). Then, we have:

\[ \frac{dt}{dx} = -\csc^2 x \implies dx = -\frac{dt}{\csc^2 x} = -\frac{dt}{1 + t^2} \]

Now, we also know that:

\[ \sin^2 x = \frac{1}{1 + t^2} \quad \text{and} \quad \cos^2 x = \frac{t^2}{1 + t^2} \]

Substituting these into the integral gives:

\[ \int \frac{6 \cdot \frac{1}{1 + t^2} \cdot \left(-\frac{dt}{1 + t^2}\right)}{\frac{1}{1 + t^2} - \frac{t^2}{1 + t^2}} = \int \frac{-6 \, dt}{1 - t^2} \]

### Step 3: Solve the integral
Now we have:

\[ -6 \int \frac{dt}{1 - t^2} \]

This integral can be solved using the formula:

\[ \int \frac{dt}{1 - t^2} = \frac{1}{2} \log \left| \frac{1 + t}{1 - t} \right| + C \]

Thus, we get:

\[ -6 \cdot \frac{1}{2} \log \left| \frac{1 + t}{1 - t} \right| + C = -3 \log \left| \frac{1 + t}{1 - t} \right| + C \]

### Step 4: Substitute back for \( t \)
Recall that \( t = \cot x \). Therefore, we substitute back:

\[ -3 \log \left| \frac{1 + \cot x}{1 - \cot x} \right| + C \]

### Final Answer
The integral evaluates to:

\[ \int \frac{6 \, dx}{\sin^2 x (1 - \cot^2 x)} = -3 \log \left| \frac{1 + \cot x}{1 - \cot x} \right| + C \]