Question

Solve : ${\int }_0^{\pi /4}\ln (1+\tan x)dx$

Answer 1

Let $I={\int }_0^{\pi /4}\ln (1+\tan x)dx$ ....$\left(1\right)$

Replace $x\to \frac{\pi }{4}-x$

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

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

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

$I={\int }_0^{\pi /4}\ln \left(\frac{2}{1-\tan x}\right)dx$

$I={\int }_0^{\pi /4}\ln \left(\frac{2dx}{1-\tan x}\right)$ .......$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$2I={\int }_0^{\pi /4}\ln [(1+\tan x)\times \left(\frac{2}{1+\tan x}\right)]dx$ using the properties of logarithm

$\Rightarrow 2I={\int }_0^{\pi /4}\ln 2dx$

$\Rightarrow 2I=\ln 2{\int }_0^{\pi /4}dx$

$\Rightarrow 2I=\ln 2[x{]}_0^{\pi /4}$

$\Rightarrow 2I=\ln 2[\frac{\pi }{4}-0]$