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

Options

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

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\cot x) \, dx \), we can use a property of definite integrals. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\cot x) \, dx \]

### Step 2: Use the Property of Integrals
We can use the property of integrals which states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \frac{\pi}{2} \). Thus, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2(\frac{\pi}{2} - x)) \log(\cot(\frac{\pi}{2} - x)) \, dx \]

### Step 3: Simplify the Integral
Now, we simplify the expressions:
- \( \sin(2(\frac{\pi}{2} - x)) = \sin(\pi - 2x) = \sin(2x) \)
- \( \cot(\frac{\pi}{2} - x) = \tan(x) \)

Thus, we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \]

### Step 4: Combine the Two Integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\cot x) \, dx \)
2. \( I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \)

Adding these two equations gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \left( \log(\cot x) + \log(\tan x) \right) \, dx \]

### Step 5: Simplify the Logarithmic Expression
Using the property of logarithms:
\[ \log(a) + \log(b) = \log(ab) \]
we can simplify:
\[ \log(\cot x) + \log(\tan x) = \log(\cot x \cdot \tan x) = \log(1) = 0 \]

### Step 6: Evaluate the Integral
Thus, we have:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \cdot 0 \, dx = 0 \]
This implies:
\[ I = 0 \]

### Final Answer
The value of the integral is:
\[ \boxed{0} \]