Question

Evaluate: ${\int }_0^{\pi /4}\log (1+\tan \theta )d\theta$

Answer 1

Let $I={\int }_0^{\frac{\pi }{4}}\ln (1+\tan \theta )d\theta$

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

$I={\int }_{\frac{\pi }{4}}^0\ln (1+\tan (\frac{\pi }{4}-\theta ))d\theta$

$I={\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1-\tan \theta }{1+\tan \theta })d\theta$

$I={\int }_0^{\frac{\pi }{4}}\ln \left(\frac{1+\tan \theta +1-\tan \theta }{1+\tan \theta }\right)d\theta$

$I={\int }_0^{\frac{\pi }{4}}\left(\ln \right(2)-\ln (1+\tan \theta \left)\right)d\theta$

$I={\int }_0^{\frac{\pi }{4}}\ln 2d\theta -I$

$2I=\frac{\pi }{4}\ln 2\Rightarrow I=\frac{\pi }{8}\ln 2$