Question

Evaluate : $\underset{0}{\overset{\pi /4}{\int }}\log (1+\tan x)dx$

Answer 1

$I=\underset{0}{\overset{\pi /4}{\int }}\log (1+\tan x)dx$ .......(1)

$I=\underset{0}{\overset{\pi /4}{\int }}\log (1+\tan (\frac{\pi }{4}-x))dx$ $(\because \underset{0}{\overset{a}{\int }}f(x)dx=\underset{0}{\overset{a}{\int }}f(a-x\left)dx\right)$

$I=\underset{0}{\overset{\pi /4}{\int }}\log [1+\frac{1-\tan x}{1+\tan x}]dx$

$I=\underset{0}{\overset{\pi /4}{\int }}\log \frac{2}{1+\tan x}dx$

$I=\underset{0}{\overset{\pi /4}{\int }}\log 2dx-\underset{0}{\overset{\pi /4}{\int }}\log (1+\tan x)dx$

$2I=\underset{0}{\overset{\pi /4}{\int }}\log 2dx=\log 2\underset{0}{\overset{\pi /4}{\int }}dx$

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

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