A) \[\frac{\pi }{8}{{\log }_{e}}2\]
B) \[\frac{\pi }{4}{{\log }_{2}}e\]
C) \[\frac{\pi }{4}{{\log }_{e}}2\]
D) \[\frac{\pi }{4}{{\log }_{e}}\left( \frac{1}{2} \right)\]
Correct Answer: A
Solution :
Let \[I=\int_{0}^{\pi /4}{\log \,\,\left[ 1+\tan \left( \frac{\pi }{4}-x \right) \right]}\,dx\] ... (i) \[\Rightarrow \] \[I=\int_{0}^{\pi /4}{\log \left[ 1+\tan \left( \frac{\pi }{4}-x \right) \right]\,\,dx}\] \[\left[ \because \,\int_{0}^{a}{f(x)\,dx=\int_{0}^{a}{f(a-x)\,dx}} \right]\] \[=\int_{0}^{\pi /4}{\log \left[ 1+\frac{1-\tan x}{1+\tan x} \right]dx}\] \[=\int_{0}^{\pi /4}{\log \left[ 1+\frac{2}{1+\tan x} \right]dx}\] \[=\int_{0}^{\pi /4}{\log \,2\,dx-\int_{0}^{\pi /4}{\log \,(1+\tan x)\,dx}}\] \[\Rightarrow \] \[I=\log \,2\,[x]_{0}^{\pi /4}-I\] [from equation (i)] \[\Rightarrow \,\,\,\,2I=\frac{\pi }{4}{{\log }_{e}}2\] \[\Rightarrow \,\,\,I=\frac{\pi }{8}\,{{\log }_{e}}2\]
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