Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$ .............. (1)
$\therefore I={\int }_0^{\frac{\pi }{4}}\log [1+\tan (\frac{\pi }{4}-x)]dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log \{1+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}\}dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log \{1+\frac{1-\tan x}{1+\tan x}\}dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log \frac{2}{(1+\tan x)}dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log 2dx-{\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log 2dx-I$ [From (1)]
$\Rightarrow 2I=[x\log 2{]}_0^{\frac{\pi }{4}}$
$\Rightarrow 2I=\frac{\pi }{4}\log 2\Rightarrow I=\frac{\pi }{8}\log 2$