Question

The value of definite integral \( \int_0^{\pi/2} \log(\tan x) dx \) is:

• A. 0
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( \pi \)

Hint:

When solving integrals of logarithmic functions, try applying symmetry and properties of the functions to simplify the process.

Answer 1

The Correct Option is A

Step 1: Let: \[ I = \int_0^{\pi/2} \log(\tan x) dx \] Apply the property: \[ \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \] Thus: \[ I = \int_0^{\pi/2} \log(\tan(\frac{\pi}{2} - x)) dx = \int_0^{\pi/2} \log(\cot x) dx \] Step 2: Adding the two equations gives: \[ 2I = \int_0^{\pi/2} [\log(\tan x) + \log(\cot x)] dx = \int_0^{\pi/2} \log(1) dx = 0 \] Therefore, \( I = 0 \).

Answer 2

Step 1: Analyze the given integral
We are asked to evaluate the integral: \[ I = \int_0^{\pi/2} \log(\tan x) \, dx \]
Step 2: Symmetry property of the integral
One approach is to use the symmetry of the tangent function. Notice that: \[ \tan\left(\frac{\pi}{2} - x\right) = \cot x \] Therefore, we can substitute \( x \) by \( \frac{\pi}{2} - x \) in the integral: \[ I = \int_0^{\pi/2} \log\left(\tan\left(\frac{\pi}{2} - x\right)\right) dx = \int_0^{\pi/2} \log(\cot x) \, dx \]
Step 3: Simplify the integral
Using the logarithmic identity \( \log(\cot x) = \log\left(\frac{1}{\tan x}\right) = -\log(\tan x) \), we can write: \[ I = \int_0^{\pi/2} \log(\cot x) \, dx = -\int_0^{\pi/2} \log(\tan x) \, dx \] Thus, we have: \[ I = -I \]
Step 4: Solve the equation
This implies that: \[ 2I = 0 \quad \Rightarrow \quad I = 0 \] Thus, the value of the definite integral is: \[ \boxed{0} \]