🧠 Quick Explanation:

Use the property of definite integrals: \[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \] Let \[ I = \int_0^{\frac{\pi}{2}} \log(\tan x) \, dx \] Then: \[ I = \int_0^{\frac{\pi}{2}} \log(\tan(\frac{\pi}{2} - x)) \, dx = \int_0^{\frac{\pi}{2}} \log(\cot x) \, dx \] Now add both: \[ 2I = \int_0^{\frac{\pi}{2}} [\log(\tan x) + \log(\cot x)] \, dx = \int_0^{\frac{\pi}{2}} \log(\tan x \cdot \cot x) \, dx \] But \(\tan x \cdot \cot x = 1\), so: \[ 2I = \int_0^{\frac{\pi}{2}} \log(1) \, dx = \int_0^{\frac{\pi}{2}} 0 \, dx = 0 \Rightarrow I = 0 \]