Question

The value of the definite integral $\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln (1+\tan x)dx$ is:

• A. $\frac{\pi }{4}\ln 2$
• B. $\frac{\pi }{8}\ln 2$
• C. $\pi \ln 2$
• D. $\ln 2$

Answer 1

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

Let $I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln (1+\tan x)dx\cdots \left(i\right)$
Using the property:
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln [1+\tan (\frac{\pi }{4}-x)]dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln [1+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}]dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln [1+\frac{1-\tan x}{1+\tan x}]dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln \frac{2}{(1+\tan x)}dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln 2dx-\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln (1+\tan x)dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\ln 2dx-I$ [From$\cdots \left(i\right)$]
$\Rightarrow 2I=\left(\ln 2\right)[x{]}_0^{\frac{\pi }{4}}$
$\Rightarrow 2I=\frac{\pi }{4}\ln 2$
$\Rightarrow I=\frac{\pi }{8}\ln 2$