If ∫ 0 π log ( sin x ) d x = 8 k , then ∫ 0 π / 4 log ( 1 + tan x ) d x =

Let $I={\int }_0^{\pi /4}\log (1+\tan x)dx$

Using property of integration ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$\Rightarrow I={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\right)dx$

$\Rightarrow I={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\frac{2}{1+\tan x}\right)dx$

$\Rightarrow I={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(2\right)dx-{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan x\right)dx$

$\Rightarrow I=\frac{\mathrm{\pi }}{4}\log \left(2\right)-I \left[\because I={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan x\right)dx\right]$

$\Rightarrow 2I=\frac{\mathrm{\pi }}{4}\log \left(2\right)$

$\Rightarrow I=\frac{\mathrm{\pi }}{8}\log \left(2\right) \left[1\right]$

$\Rightarrow {\int }_0^{\mathrm{\pi }}\log \left(\sin x\right)dx=8k$

$\Rightarrow -\mathrm{\pi } \log \left(2\right)=8k \left[\text{from equation }1\right]$

$\Rightarrow -8I=8k$

$\Rightarrow I=-k$

$\Rightarrow {\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan x\right) =-k$