Question

Show that: ${\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx=\frac{\pi }{8}\log 2$

Answer 1

Consider the given integral.

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

On applying property

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Therefore,

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

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

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

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

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

$I={\int }_0^{\pi /4}[\log 2-\log (1+\tan x\left)\right]dx$ $............\left(2\right)$

On adding eqaution $\left(1\right)$ and $\left(2\right)$, we get

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

$2I=\log 2{\left[x\right]}_0^{\pi /4}$

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

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

Hence, this is the answer.