Question

Integrate:${\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

Answer 1

${\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

Since ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$={\int }_0^{\frac{\pi }{4}}\log (1+\tan (\frac{\pi }{4}-x))dx$

$={\int }_0^{\frac{\pi }{4}}\log (1+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x})dx$

$={\int }_0^{\frac{\pi }{4}}\log (1+\frac{1-\tan x}{1+\tan x})dx$

$={\int }_0^{\frac{\pi }{4}}\log \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right)dx$

$={\int }_0^{\frac{\pi }{4}}\log \left(\frac{2}{1+\tan x}\right)dx$

$={\int }_0^{\frac{\pi }{4}}\log 2dx-{\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

$\Rightarrow I=\log 2{\left[x\right]}_0^{\frac{\pi }{4}}-I$ since $I={\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

$\Rightarrow 2I=\log 2\times [\frac{\pi }{4}-0]$

$\Rightarrow I=\frac{\pi \log 2}{8}$