Question
Evaluate: $\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$

Answer

$\text { Let } I =\int_0^\pi \frac{1}{3+2 \sin x+\cos x} d x$
Put $\tan \left(\frac{x}{2}\right)= t$
$\therefore x =2 \tan ^{-1} t$
$\therefore dx =\frac{2 dt }{1+ t ^2}, \sin x =\frac{2 t }{1+ t ^2}$ and $\cos x =\frac{1- t ^2}{1+ t ^2}$
When $x =0, t =0$ and when $x =\pi, t =\infty$
$ \therefore I =\int_0^{\infty} \frac{1}{3+2\left(\frac{2 t }{1+ t ^2}\right)+\frac{1- t ^2}{1+ t ^2}} \times \frac{2 dt }{1+ t ^2}$
$=\int_0^{\infty} \frac{2 dt }{3+3 t ^2+4 t +1- t ^2}$
$=\int_0^{\infty} \frac{2 dt }{2 t ^2+4 t +4}$
$=\int_0^{\infty} \frac{ dt }{ t ^2+2 t +2} $
$=\int_0^{\infty} \frac{ dt }{ t ^2+2 t +1+1}$
$=\int_0^{\infty} \frac{ dt }{( t +1)^2+1^2}$
$=\left[\tan ^{-1}( t +1)\right]_0^{\infty}$
$=\tan ^{-1}(1+\infty)-\tan ^{-1}(1+0)$
$=\tan ^{-1}(\infty)-\tan ^{-1}(1)$
$=\frac{\pi}{2}-\frac{\pi}{4}$
$\therefore I ==\frac{\pi}{4}$

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