Question

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

• A. $\frac{\pi }{8}lo{g}_{e}2$
• B. $\frac{\pi }{4}lo{g}_{e}2$
• C. $\frac{\pi }{4}$
• D. none of these

Answer 1

Answer: (A)

$\frac{\pi }{8}lo{g}_{e}2$

$I={\int }_0^{\pi /4}log(1+\tan x)dx={\int }_0^{\pi /4}\log (1+\tan (\frac{\pi }{4}-x))dx$
$I={\int }_0^{\pi /4}log(1+\tan x)dx={\int }_0^{\pi /4}\log (1+\tan (\frac{\pi }{4}-x))dx$
$={\int }_0^{\pi /4}\log 2dx-{\int }_0^{\pi /4}log(1+\tan x)dx=\frac{\pi }{4}\log 2-I$
$\Rightarrow I=\frac{\pi }{8}\log 2$