Question

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

Answer 1

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

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

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

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

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

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

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

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