What is the value of `int_(0)^(pi/2) log (tanx) dx` ?

Options

A. 0
B. 1
C. `-1`
D. `pi//4`

Explanation

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

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

### 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, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\tan(\frac{\pi}{2} - x)) \, dx. \]

### Step 3: Simplify the Integral
Using the identity \( \tan(\frac{\pi}{2} - x) = \cot x \), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\cot x) \, dx. \]
Since \( \cot x = \frac{1}{\tan x} \), we have:
\[ \log(\cot x) = \log\left(\frac{1}{\tan x}\right) = -\log(\tan x). \]
Thus, we can express \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{2}} -\log(\tan x) \, dx = -I. \]

### Step 4: Solve for \( I \)
Now, we can add \( I \) to both sides:
\[ I + I = 0 \implies 2I = 0 \implies I = 0. \]

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