Question

${\int }_0^{\pi /4}\log (1+\tan x)dx$ is equal to

• A. $\frac{\pi }{8}{\log }_{e}2$
• B. $\frac{\pi }{4}{\log }_{e}e$
• C. $\frac{\pi }{4}{\log }_{e}2$
• D. $\frac{\pi }{8}{\log }_e\left(\frac{1}{2}\right)$

Answer 1

The correct option is A $\frac{\pi }{8}{\log }_{e}2$

Let $I={\int }_0^{\pi /4}\log (1+\tan x)dx.....\left(i\right)$
$\Rightarrow I={\int }_0^{\pi /4}\log [1+\tan (\frac{\pi }{4}-x)]dx$
$[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]$
$={\int }_0^{\pi /4}\log [1+\frac{1-\tan x}{1+\tan x}]dx$
$={\int }_0^{\pi /4}\log \left[\frac{2}{1+\tan x}\right]dx$$
$={\int }_0^{\pi /4}\log 2dx-{\int }_0^{\pi /4}\log (1+\tan x)dx$
$\Rightarrow I=\log 2[x{]}_0^{\pi /4}-I$ [from Eq. (i)]
$\Rightarrow 2I=\frac{\pi }{4}{\log }_{e}2$
$\Rightarrow I=\frac{\pi }{8}{\log }_{e}2$