Evaluate : `int_(0)^((pi)/(2)) log (cot x)dx`.

Answer

To evaluate the integral \( I = \int_0^{\frac{\pi}{2}} \log(\cot x) \, dx \), we can follow these steps:

### Step 1: Define the Integral
Let
\[ I = \int_0^{\frac{\pi}{2}} \log(\cot x) \, dx \]

### Step 2: Use the Property of Integrals
We can use the property of integrals that 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,
\[ I = \int_0^{\frac{\pi}{2}} \log(\cot(\frac{\pi}{2} - x)) \, dx \]

### Step 3: Simplify the Argument
We know that:
\[ \cot\left(\frac{\pi}{2} - x\right) = \tan x \]
So we can rewrite the integral as:
\[ I = \int_0^{\frac{\pi}{2}} \log(\tan x) \, dx \]

### Step 4: Add the Two Integrals
Now we have two expressions for \( I \):
\[ I = \int_0^{\frac{\pi}{2}} \log(\cot x) \, dx \]
\[ I = \int_0^{\frac{\pi}{2}} \log(\tan x) \, dx \]
Adding these two equations gives:
\[ 2I = \int_0^{\frac{\pi}{2}} \left( \log(\cot x) + \log(\tan x) \right) \, dx \]

### Step 5: Use Logarithmic Properties
Using the property of logarithms, we can combine the logs:
\[ \log(\cot x) + \log(\tan x) = \log(\cot x \cdot \tan x) = \log(1) \]
Since \( \cot x \cdot \tan x = 1 \).

### Step 6: Evaluate the Integral
Thus, we have:
\[ 2I = \int_0^{\frac{\pi}{2}} \log(1) \, dx \]
Since \( \log(1) = 0 \):
\[ 2I = \int_0^{\frac{\pi}{2}} 0 \, dx = 0 \]

### Step 7: Solve for \( I \)
Therefore, we find:
\[ 2I = 0 \implies I = 0 \]

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