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

Options

A. `ln 2`
B. `-ln 2`
C. `0`
D. None of these

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \ln(\tan x) \, dx \), we can use a symmetry property of definite integrals. Let's follow the steps:

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

### Step 2: Use the property of logarithms
We can express \(\tan x\) in terms of \(\cot x\):
\[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \]
Thus,
\[ \ln(\tan x) = \ln\left(\frac{\sin x}{\cos x}\right) = \ln(\sin x) - \ln(\cos x) \]

### Step 3: Change the variable
Now, we will change the variable in the integral:
Let \( u = \frac{\pi}{2} - x \). Then, \( du = -dx \) and the limits change as follows:
- When \( x = 0 \), \( u = \frac{\pi}{2} \)
- When \( x = \frac{\pi}{2} \), \( u = 0 \)

Thus, we can rewrite the integral:
\[ I = \int_{\frac{\pi}{2}}^{0} \ln(\tan(\frac{\pi}{2} - u)) (-du) = \int_{0}^{\frac{\pi}{2}} \ln(\cot u) \, du \]

### Step 4: Simplify the integral
Using the identity \(\cot u = \frac{1}{\tan u}\), we have:
\[ \ln(\cot u) = \ln\left(\frac{1}{\tan u}\right) = -\ln(\tan u) \]
So,
\[ I = \int_{0}^{\frac{\pi}{2}} -\ln(\tan u) \, du = -\int_{0}^{\frac{\pi}{2}} \ln(\tan u) \, du = -I \]

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

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