What is the value of `int_0^(pi//2) sin 2x ln (cot x)` dx

Options

A. A. 0
B. B. `pi ln 2`
C. C. `-pi ln2`
D. D. `(pi ln 2)/`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \sin(2x) \ln(\cot x) \, dx \), we can use the property of definite integrals that states:

\[ \int_0^A f(x) \, dx = \int_0^A f(A - x) \, dx \]

In our case, \( A = \frac{\pi}{2} \). Therefore, we can rewrite the integral as follows:

### Step 1: Apply the property of definite integrals
\[ I = \int_0^{\frac{\pi}{2}} \sin(2x) \ln(\cot x) \, dx = \int_0^{\frac{\pi}{2}} \sin(2(\frac{\pi}{2} - x)) \ln(\cot(\frac{\pi}{2} - x)) \, dx \]

### Step 2: Simplify the expression
Now, we simplify the terms:
- \( \sin(2(\frac{\pi}{2} - x)) = \sin(\pi - 2x) = \sin(2x) \)
- \( \cot(\frac{\pi}{2} - x) = \tan(x) \)

Thus, we have:
\[ I = \int_0^{\frac{\pi}{2}} \sin(2x) \ln(\tan x) \, dx \]

### Step 3: Combine the two integrals
Now we have two expressions for \( I \):
1. \( I = \int_0^{\frac{\pi}{2}} \sin(2x) \ln(\cot x) \, dx \)
2. \( I = \int_0^{\frac{\pi}{2}} \sin(2x) \ln(\tan x) \, dx \)

We can express the second integral in terms of the first:
\[ I = \int_0^{\frac{\pi}{2}} \sin(2x) \ln(\tan x) \, dx = \int_0^{\frac{\pi}{2}} \sin(2x) \ln\left(\frac{1}{\cot x}\right) \, dx \]

### Step 4: Rewrite the logarithmic expression
Using the property of logarithms:
\[ \ln(\tan x) = -\ln(\cot x) \]
Thus, we can write:
\[ I = \int_0^{\frac{\pi}{2}} \sin(2x) (-\ln(\cot x)) \, dx = -I \]

### Step 5: Solve for \( I \)
Adding both expressions for \( I \):
\[ I + I = 0 \implies 2I = 0 \implies I = 0 \]

### Conclusion
The value of the integral is:
\[ \int_0^{\frac{\pi}{2}} \sin(2x) \ln(\cot x) \, dx = 0 \]