A) \[\frac{\pi }{4}\log 2\]
B) \[\frac{\pi }{4}\log \frac{1}{2}\]
C) \[\frac{\pi }{8}\log 2\]
D) \[\frac{\pi }{8}\log \frac{1}{2}\]
Correct Answer: C
Solution :
\[I=\int_{0}^{\pi /4}{\,\,\,\log (1+\tan \theta )d\theta }\]Þ \[I=\int_{0}^{\pi /4}{\log \left\{ 1+\tan \left( \frac{\pi }{4}-\theta \right) \right\}}\,d\theta \] Þ I = \[\int_{0}^{\pi /4}{\log \left( 1+\frac{1-\tan \theta }{1+\tan \theta } \right)\,d\theta }\] Þ 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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