Problem Statement
Evaluate the definite integral:
$$I = \int_{0}^{\pi/4} \log(1 + \tan x) \, dx$$
Verified Solution & Marking Scheme
Apply Property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a-x)dx
Replace $x$ by $\frac{\pi}{4} - x$:
$$I = \int_{0}^{\pi/4} \log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right) dx$$
Using $\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}$:
$$1 + \tan\left(\frac{\pi}{4} - x\right) = 1 + \frac{1 - \tan x}{1 + \tan x} = \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} = \frac{2}{1 + \tan x}$$
Split the Logarithm
$$I = \int_{0}^{\pi/4} \log\left(\frac{2}{1 + \tan x}\right) dx = \int_{0}^{\pi/4} [\log 2 - \log(1 + \tan x)] dx$$
$$I = \int_{0}^{\pi/4} \log 2 \, dx - \int_{0}^{\pi/4} \log(1 + \tan x) \, dx = \log 2 [x]_{0}^{\pi/4} - I$$
Solve for I
$$2I = \frac{\pi}{4} \log 2 \implies I = \frac{\pi}{8} \log 2$$