A) \[\frac{\pi }{2}{{\log }_{e}}2\]
B) \[-\frac{\pi }{2}{{\log }_{e}}2\]
C) \[\pi {{\log }_{e}}2\]
D) 0
Correct Answer: D
Solution :
\[\int_{0}^{\pi /2}{\log \tan x\,dx=}\int_{0}^{\pi /2}{\log \left( \frac{\sin x}{\cos x} \right)dx}\] \[=\int_{0}^{\pi /2}{\log \sin x\,dx-\int_{0}^{\pi /2}{\log \cos x\,dx=0}}\], \[\left\{ \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right\}\].
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