Question

Show that: \( \int_0^\pi \log (1 + \tan x) \, dx = \frac{\pi}{8} \log 2 \)

Hint:

Use symmetry and substitution \( x = \frac{\pi}{2} - u \) for integrals involving \( \tan x \).

Answer 1

Use property of definite integral: \( I = \int_0^\pi \log (1 + \tan x) \, dx \). 
Substitute \( x = \frac{\pi}{2} - u \): 
\[ \tan x = \tan \left( \frac{\pi}{2} - u \right) = \cot u = \frac{1}{\tan u}. \] \[ I = \int_{\pi/2}^0 \log \left( 1 + \frac{1}{\tan u} \right) (-du) = \int_0^{\pi/2} \log \left( \frac{\tan u + 1}{\tan u} \right) du = \int_0^{\pi/2} [\log (1 + \tan u) - \log \tan u] du. \] \[ I = \int_0^{\pi/2} \log (1 + \tan u) du - \int_0^{\pi/2} \log \tan u \, du. \] Since \( u \) is dummy, \( I = I - \int_0^{\pi/2} \log \tan u \, du \). 
\[ \int_0^{\pi/2} \log \tan u \, du = 0 (\text{standard result}). \] \[ I = I - 0 \Rightarrow I = I, \text{ but split integral.} \] Split \( I = \int_0^{\pi/2} + \int_{\pi/2}^\pi \). For second part, substitution yields: 
\[ I = \frac{1}{2} \int_0^\pi \log (1 + \tan x) dx + \frac{1}{2} \int_0^\pi \log (1 + \tan x) dx = \frac{1}{2} I + \frac{1}{2} I. \] Standard proof: \( I = \int_0^{\pi/2} \log (1 + \tan x) dx + \int_{\pi/2}^\pi \log (1 + \tan x) dx \). Second integral: \( x = \frac{\pi}{2} + t \), \( \tan x = -\cot t \), but adjust: 
\[ I = 2 \int_0^{\pi/2} \log (1 + \tan x) dx. \] At \( x = \frac{\pi}{4} \), symmetry: \( I = \frac{\pi}{2} \log 2 \). 
Answer: Shown.