Question

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

• A. $\pi \log 2$
• B. $\frac{\pi \log 2}{2}$
• C. $\frac{\left(\pi \log 2\right)}{8}$
  
• D. $\log 2$

Answer 1

Answer: (C)

$\frac{\left(\pi \log 2\right)}{8}$

${\int }_0^{\pi /4}\log (1+\tan \theta )d\theta ...\left(1\right)$
$I={\int }_0^{\pi /4}\log (1+\tan (\frac{\pi }{4}-\theta ))d\theta ({\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx)$
$={\int }_0^{\pi /4}\log (1+\frac{\tan \left(\pi /4\right)-\tan \theta }{\tan \left(\pi /4\right)\tan \theta })d\theta$
$={\int }_0^{\pi /4}\log \left(\frac{2}{1+\tan \theta }\right)d\theta ={\int }_0^{\pi /4}[\log 2-\log (1+\tan \theta )]d\theta ...\left(2\right)$
Adding (1) and (2)
$2I={\int }_0^{\pi /4}\log (1+\tan \theta )d\theta +{\int }_0^{\pi /4}\log 2d\theta -{\int }_0^{\pi /4}\log (1+\tan \theta )d\theta$
$\Rightarrow 2I=(\frac{\pi }{4}-0)\log 2\Rightarrow I=\frac{\pi }{8}\log 2$