What is `int_0^(pi//2) ln (tan x)` equal to

Options

A. ln 2
B. `ln 2`
C. 0
D. none of these

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \ln(\tan x) \, dx \), we can use a property of definite integrals. The property states that:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

In our case, we have \( a = 0 \) and \( b = \frac{\pi}{2} \). Therefore, we can rewrite the integral as:

1. **Step 1: Apply the property of definite integrals.**

\[ I = \int_0^{\frac{\pi}{2}} \ln(\tan x) \, dx = \int_0^{\frac{\pi}{2}} \ln(\tan\left(\frac{\pi}{2} - x\right)) \, dx \]

Since \( \tan\left(\frac{\pi}{2} - x\right) = \cot x \), we can substitute this into the integral:

\[ I = \int_0^{\frac{\pi}{2}} \ln(\cot x) \, dx \]

2. **Step 2: Rewrite the integral.**

We know that \( \cot x = \frac{1}{\tan x} \), so we can express \( \ln(\cot x) \) as:

\[ \ln(\cot x) = \ln\left(\frac{1}{\tan x}\right) = -\ln(\tan x) \]

Therefore, we have:

\[ I = \int_0^{\frac{\pi}{2}} -\ln(\tan x) \, dx = -\int_0^{\frac{\pi}{2}} \ln(\tan x) \, dx = -I \]

3. **Step 3: Solve for \( I \).**

Adding \( I \) to both sides gives:

\[ I + I = 0 \implies 2I = 0 \implies I = 0 \]

Thus, the value of the integral \( I = \int_0^{\frac{\pi}{2}} \ln(\tan x) \, dx \) is equal to **0**.