The value of `int(ln(cotx))/(sin2x)dx` is equal to (where, C is the constant of integration)

Options

A. `((ln(cotx))^(2))/(2)+C`
B. `((ln(cotx))^(2))/(4)+C`
C. `((ln(cotx))^(2))/(6)+C`
D. `-(1)/(4)(ln(cotx))^(2)+C`

Answer

To solve the integral \( I = \int \frac{\ln(\cot x)}{\sin(2x)} \, dx \), we will use substitution and integration techniques. Here’s the step-by-step solution:

### Step 1: Set up the integral
Let \( I = \int \frac{\ln(\cot x)}{\sin(2x)} \, dx \).

### Step 2: Use substitution for \( \ln(\cot x) \)
We will substitute \( t = \ln(\cot x) \). To find \( dt \), we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = \frac{d}{dx} \ln(\cot x) = \frac{1}{\cot x} \cdot \frac{d}{dx}(\cot x) \]
The derivative of \( \cot x \) is \( -\csc^2 x \), so:
\[ \frac{dt}{dx} = \frac{-\csc^2 x}{\cot x} = -\frac{1}{\sin^2 x} \cdot \frac{1}{\cot x} = -\frac{1}{\sin^2 x} \cdot \frac{\sin x}{\cos x} = -\frac{1}{\sin x \cos x} \]

### Step 3: Express \( dx \) in terms of \( dt \)
Rearranging gives:
\[ dx = -\sin x \cos x \, dt \]

### Step 4: Substitute in the integral
Now, we need to express \( \sin(2x) \) in terms of \( t \):
\[ \sin(2x) = 2\sin x \cos x \]
Thus, the integral becomes:
\[ I = \int \frac{t}{\sin(2x)} \cdot (-\sin x \cos x) \, dt = -\int \frac{t}{2\sin x \cos x} \cdot \sin x \cos x \, dt = -\frac{1}{2} \int t \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ -\frac{1}{2} \int t \, dt = -\frac{1}{2} \cdot \frac{t^2}{2} + C = -\frac{1}{4} t^2 + C \]

### Step 6: Substitute back for \( t \)
Recall that \( t = \ln(\cot x) \):
\[ I = -\frac{1}{4} (\ln(\cot x))^2 + C \]

### Final Result
Thus, the value of the integral is:
\[ I = -\frac{1}{4} (\ln(\cot x))^2 + C \]