`int_(0)^(pi//2) log (cotx ) dx=`

Options

A. `pi/4`
B. `pi/2`
C. 0
D. `pi`

Answer

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

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

### Step 2: Use the Property of Definite Integrals
We can use the property of definite 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, we can write:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\cot(\frac{\pi}{2} - x)) \, dx \]

### Step 3: Simplify the Integral
Using the trigonometric identity \( \cot(\frac{\pi}{2} - x) = \tan x \), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx \]

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

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

### Step 5: Use Logarithmic Properties
Using the property of logarithms, \( \log(a) + \log(b) = \log(ab) \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(\cot x \cdot \tan x) \, dx \]
Since \( \cot x \cdot \tan x = 1 \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(1) \, dx \]

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

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

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