Question

Prove that $\stackrel{\pi /4}{\underset{0}{\int }}log(1+\tan x)dx=\frac{\pi }{8}lo{g}_e^2$.

Answer 1

$I=\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log(1+\tan x)dx$
$\to I=\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log(1+tanx)dx$
$=\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log[1+\tan (\frac{\pi }{4}-x)]dx$
$(\because \stackrel{a}{\underset{0}{\int }}f\left(x\right)dx=\stackrel{a}{\underset{0}{\int }}f(a-x)dx)$
$=\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log[1+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\cdot \tan x}]dx$
$=\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log[1+\frac{1-\tan x}{1+\tan x}]dx$
$=\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log\left[\frac{1+\tan x+1-\tan x}{1+\tan x}\right]dx$
$=\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log\left(\frac{2}{1+\tan x}\right)dx$
$=log2dx-\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log(1+\tan x)dx$
$\therefore =\stackrel{\frac{\pi }{4}}{\underset{0}{\int }}log2dx-I$
$\therefore 2I=log2[x{]}_0^{\pi /4}$
$=log2(\frac{\pi }{4}-0)=\frac{\pi }{4}\cdot lo{g}_{e}2$
$\therefore I=\frac{\pi }{8}log2$.