If `int ("In"(cotx))/(sinx cos x) dx=(-1)/(k)"In"^(2)(cotx)+C`
(where C is a constant), then the value of k is :

Options

A. 1
B. 2
C. 3
D. `(1)/(2)`

Answer

To solve the given problem, we need to evaluate the integral and find the value of \( k \) such that:

\[ \int \frac{\ln(\cot x)}{\sin x \cos x} \, dx = -\frac{1}{k} \ln^2(\cot x) + C \]

### Step-by-step Solution:

1. **Differentiate Both Sides:**
Start by differentiating both sides of the equation with respect to \( x \).

\[ \frac{d}{dx} \left( \int \frac{\ln(\cot x)}{\sin x \cos x} \, dx \right) = \frac{d}{dx} \left( -\frac{1}{k} \ln^2(\cot x) + C \right) \]

2. **Apply the Fundamental Theorem of Calculus:**
By the Fundamental Theorem of Calculus, the left side simplifies to:

\[ \frac{\ln(\cot x)}{\sin x \cos x} \]

3. **Differentiate the Right Side:**
Now differentiate the right side. Using the chain rule:

\[ \frac{d}{dx} \left( -\frac{1}{k} \ln^2(\cot x) \right) = -\frac{2}{k} \ln(\cot x) \cdot \frac{d}{dx}(\ln(\cot x)) \]

The derivative of \( \ln(\cot x) \) is:

\[ \frac{d}{dx}(\ln(\cot x)) = \frac{-\csc^2(x)}{\cot x} = -\frac{1}{\sin x \cos x} \]

Therefore, the derivative becomes:

\[ -\frac{2}{k} \ln(\cot x) \cdot \left(-\frac{1}{\sin x \cos x}\right) = \frac{2 \ln(\cot x)}{k \sin x \cos x} \]

4. **Set the Derivatives Equal:**
Now we have:

\[ \frac{\ln(\cot x)}{\sin x \cos x} = \frac{2 \ln(\cot x)}{k \sin x \cos x} \]

5. **Cancel Common Terms:**
Assuming \( \ln(\cot x) \neq 0 \), we can cancel \( \ln(\cot x) \) from both sides:

\[ 1 = \frac{2}{k} \]

6. **Solve for \( k \):**
Rearranging gives:

\[ k = 2 \]

### Final Answer:
Thus, the value of \( k \) is \( 2 \).

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