Question

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

Answer 1

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

= ${\displaystyle {\int }_0^{\frac{\pi }{4}}\log \{1+\tan (\frac{\pi }{4}-x)\}dx}$   ......${\displaystyle (\because {\int }_0^{a}f\left(x\right)dx=\int f(a -x)dx)}$

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

= ${\displaystyle {\int }_0^{\frac{\pi }{4}}\{1+\frac{1-\tan x}{1+\tan x}\}dx}$

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

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

= ${\displaystyle {\int }_0^{\frac{\pi }{4}}\{\log 2-\log (1+\tan x)\}dx}$

= ${\displaystyle {\int }_0^{\frac{\pi }{4}}\log 2 dx -{\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx}$

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

2I = ${\displaystyle \log 2[\frac{\pi }{4}-0]}$

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

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