Question

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

• A. $\left(\pi /8\right)\log 2.$
• B. $\left(\pi /16\right)\log 4.$
• C. $\left(\pi /8\right)\log \left(1/2\right).$
• D. $(-\pi /8)\log 2.$

Answer 1

Answer: (A), (B)

The correct options are
A $\left(\pi /8\right)\log 2.$
B $\left(\pi /16\right)\log 4.$

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

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

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

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

$\Rightarrow 2I=\frac{\pi }{4}\log 2=\frac{\pi }{8}\log 2$

$\Rightarrow I=\frac{\pi }{16}\log 4$