Integrals: Definite Integral - Logarithmic Trigonometric Integral

Let equation (1) be: $$I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \quad \text{--- (1)}$$ Using $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$ with $a = \frac{\pi}{4}$: $$I = \int_{0}^{\frac{\pi}{4}} \log\left[1 + \tan\left(\frac{\pi}{4} - x\right)\right] \, dx$$

Since $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$: $$\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}$$ $$1 + \tan\left(\frac{\pi}{4} - x\right) = 1 + \frac{1 - \tan x}{1 + \tan x} = \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} = \frac{2}{1 + \tan x}$$ Therefore: $$I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan x}\right) \, dx = \int_{0}^{\frac{\pi}{4}} [\log 2 - \log(1 + \tan x)] \, dx \quad \text{--- (2)}$$

$$I + I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx + \int_{0}^{\frac{\pi}{4}} \log 2 \, dx - \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx$$ $$2I = \log 2 \int_{0}^{\frac{\pi}{4}} dx = \log 2 [x]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} \log 2$$ $$I = \frac{\pi}{8} \log 2$$