Integrals — Class 12 Maths Solution

Let$I = \int\limits_0^{\pi /4} {\log \left( {1 + \tan x} \right)dx}$

…(i)
Also, $I = \int\limits_0^{\pi /4} {\log \left[ {1 + \tan \left( {\cfrac{\pi }{4} - x} \right)} \right]} dx$

$\Rightarrow$ $I = \int\limits_0^{\pi /4} {\log \left( {1 + \cfrac{{1 - \tan x}}{{1 + \tan x}}} \right)} dx$

$= \int\limits_0^{\pi /4} {\log \left( {\cfrac{2}{{1 + \tan x}}} \right)} dx$

$= \int\limits_0^{\pi /4} {\log 2dx} - \int\limits_0^{\pi /4} {\log \left( {1 + \tan x} \right)dx} = \log 2\int\limits_0^{\pi /4} {1\,dx} - I$

$= 2I = \log 2\left[ x \right]_0^{\pi /4} = \left( {\log 2} \right)\left( {\cfrac{\pi }{4} - 0} \right)$ $\Rightarrow$ $I = \cfrac{\pi }{8}\log 2$