Let \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + tan x }\)
⇒ \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + tan (\frac {\pi}4- x) }\)
\(\left(\because \int\limits_a^b f(x) dx = \int\limits^b_af(a + b - x)dx\right)\)
⇒ \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + \cfrac{tan\frac{\pi}4 - tanx}{1 + tan\frac{\pi}{4}tanx} }\)
\(= \int\limits_{0}^{\frac{\pi}4} \frac{dx}{ \cfrac{2 + tanx + 1 - tanx}{1 + tanx}}\) \(\left(\because tan \frac{\pi}{4} = 1\right)\)
\(= \int\limits_{0}^{\frac{\pi}4} \frac{1 + tan x }{2}dx\)
\(= \frac12 [x + log\, secx] ^{\frac{\pi}4}_0\) \((\because \int tan x \,dx = log\,secx)\)
\(= \frac 12 \left[\left(\frac{\pi}{4} - 0\right) + log \left(sec\frac{\pi}{4}\right) - log\,sec 0\right]\)
\(= \frac12 \left(\frac{\pi}{4} + log \sqrt 2 - log \,1\right)\)
\(= \frac 12 \left( \frac{\pi}{4} + \frac12 log2\right)\) \((\because log 1 = 0)\)
\(= \frac{\pi}6 + \frac14 log2.\)
