Question
Show that : $\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x=\frac{\pi}{8} \log 2$
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Answer
$\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x=\frac{\pi}{8} \log 2$
$\text { I }=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$
$=\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x$
$\ldots\left(\int_0^a f(x) d x=\int_0^a f(a-x) d x\right)$
$\therefore I=\int_0^{\frac{\pi}{4}} \log \left[\frac{1+\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \cdot \tan x}\right] d x$
$=\int_0^{\frac{\pi}{4}} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x \ldots\left(\because \tan \frac{\pi}{4}=1\right)$
$=\int_0^{\frac{\pi}{4}} \log \left[\frac{1+\tan x+1-\tan x}{1+\tan x}\right] d x$
$I=\int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x....(2)$
Adding $(1)$ and $(2)$
$2 I =\int_0^{\frac{\pi}{4}}\left[\log (1+\tan x)+\log \left(\frac{2}{1+\tan x}\right)\right] d x$
$ =\int_0^{\frac{\pi}{4}} \log 2 d x$
$ =\log 2 \int_0^{\frac{\pi}{4}} 1 d x$
$ =[x \log 2]_0^{\pi / 4}$
$ =\frac{\pi}{4} \log 2$
$\therefore I =\frac{\pi}{8} \log 2=\text { R.H.S. }$
$\text { I }=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$
$=\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x$
$\ldots\left(\int_0^a f(x) d x=\int_0^a f(a-x) d x\right)$
$\therefore I=\int_0^{\frac{\pi}{4}} \log \left[\frac{1+\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \cdot \tan x}\right] d x$
$=\int_0^{\frac{\pi}{4}} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x \ldots\left(\because \tan \frac{\pi}{4}=1\right)$
$=\int_0^{\frac{\pi}{4}} \log \left[\frac{1+\tan x+1-\tan x}{1+\tan x}\right] d x$
$I=\int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x....(2)$
Adding $(1)$ and $(2)$
$2 I =\int_0^{\frac{\pi}{4}}\left[\log (1+\tan x)+\log \left(\frac{2}{1+\tan x}\right)\right] d x$
$ =\int_0^{\frac{\pi}{4}} \log 2 d x$
$ =\log 2 \int_0^{\frac{\pi}{4}} 1 d x$
$ =[x \log 2]_0^{\pi / 4}$
$ =\frac{\pi}{4} \log 2$
$\therefore I =\frac{\pi}{8} \log 2=\text { R.H.S. }$
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