Question

The value of $\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+\tan x)dx$ is

• A. $\frac{\pi }{4}-\log 2$
• B. $\frac{\pi }{8}\log 2$
• C. $\frac{\pi }{8}+\log 2$
• D. $\frac{\pi }{4}+\log 4$

Answer 1

The correct option is B $\frac{\pi }{8}\log 2$

$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+\tan x)dx$
Using property of integration
$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+\tan (\frac{\pi }{4}-x))dx$
$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x})dx$
$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+\frac{1-\tan x}{1+\tan x})dxI=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log \left(\frac{2}{1+\tan x}\right)dxI=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log 2dx-\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+\tan x)dx$
$I=\frac{\pi }{4}\log 2-I\Rightarrow 2I=\frac{\pi }{4}\log 2$
$I=\frac{\pi }{8}\log 2$