Question
$\int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta = } $
==> $I = \int_0^{\pi /4} {\log \left\{ {1 + \tan \left( {\frac{\pi }{4} - \theta } \right)} \right\}} \,d\theta $
==> I = $\int_0^{\pi /4} {\log \left( {1 + \frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)\,d\theta } $
==> I = $\int_0^{\pi /4} {\log 2d\theta - \int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta } } $
$ \Rightarrow I = \frac{1}{2}\int_0^{\pi /4} {\log 2d\theta = \frac{{\log 2}}{2}|\theta |_0^{\pi /4} = \frac{\pi }{8}\log 2} $.
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