What is `int_(0)^(pi//2) sin 2x ln (cot x) dx` equal to ?

Options

A. `0`
B. `pi ln 2`
C. `-pi ln 2`
D. `(pi ln 2)/(2)`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \ln(\cot x) \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We know that \( \cot x = \frac{\cos x}{\sin x} \). Therefore, we can rewrite the logarithm:
\[ \ln(\cot x) = \ln(\cos x) - \ln(\sin x) \]
Thus, the integral becomes:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \left( \ln(\cos x) - \ln(\sin x) \right) \, dx \]
This can be split into two separate integrals:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \ln(\cos x) \, dx - \int_{0}^{\frac{\pi}{2}} \sin(2x) \ln(\sin x) \, dx \]

### Step 2: Evaluate the first integral
Let:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \sin(2x) \ln(\cos x) \, dx \]
And:
\[ I_2 = \int_{0}^{\frac{\pi}{2}} \sin(2x) \ln(\sin x) \, dx \]
Now, we have:
\[ I = I_1 - I_2 \]

### Step 3: Use symmetry in the integrals
To evaluate \( I_1 \), we can use the substitution \( x = \frac{\pi}{2} - u \). Then, \( dx = -du \), and the limits change from \( 0 \) to \( \frac{\pi}{2} \) to \( \frac{\pi}{2} \) to \( 0 \):
\[ I_1 = \int_{\frac{\pi}{2}}^{0} \sin(2(\frac{\pi}{2} - u)) \ln(\cos(\frac{\pi}{2} - u)) (-du) \]
This simplifies to:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \sin(2u) \ln(\sin u) \, du = I_2 \]

### Step 4: Combine the results
Since \( I_1 = I_2 \), we can substitute back into our expression for \( I \):
\[ I = I_1 - I_2 = I_2 - I_2 = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \sin(2x) \ln(\cot x) \, dx = 0 \]