<p>Let \( I = \int_0^{\pi/2} \log \tan x \, dx \). Find the value of \(I\).</p>
Step-by-Step Solution
Key Concept: Use the property that ∫₀^(π/2) f(x)dx = ∫₀^(π/2) f(π/2 - x)dx, then add the original integral to its transformed version to show the integrand equals its own negative.
<p><strong>Step 1:</strong> Let I = ∫₀^(π/2) log(tan x) dx</p><p><strong>Step 2:</strong> Apply the substitution property: Replace x with (π/2 - x), so dx = -dx</p><p>I = ∫_(π/2)^0 log(tan(π/2 - x)) · (-dx) = ∫₀^(π/2) log(cot x) dx</p><p><strong>Step 3:</strong> Use the identity log(cot x) = log(1/tan x) = -log(tan x)</p><p>I = ∫₀^(π/2) (-log(tan x)) dx = -∫₀^(π/2) log(tan x) dx = -I</p><p><strong>Step 4:</strong> From I = -I, we get 2I = 0</p><p>∴ Answer: <strong>I = 0</strong></p>
Correct Answer: 0