A) \[\frac{\pi }{4}\log 2\]
B) \[\frac{\pi }{8}\log 2\]
C) \[\frac{\pi }{4}\log \tan x\]
D) zero
Correct Answer: B
Solution :
\[I=\int_{0}^{\pi /4}{\log (1+\tan x)\,dx}\] ?(i) \[=\int_{0}^{\pi /4}{\log (1+\tan (\pi /4-x)}\,dx\] \[I=\int_{0}^{\pi /4}{\log \left( 1+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x} \right)}\,dx\] \[I=\int_{0}^{\pi /4}{\log \left( 1+\frac{1-\tan x}{1+\tan x} \right)dx}\] \[I=\int_{0}^{\pi /4}{\log \left( \frac{2}{1+\tan x} \right)\,}dx\] \[I=\int_{0}^{\pi /4}{\log 2dx-\int_{0}^{\pi /3}{\log (1+\tan x)dx}}\] \[2I=\log 2\left[ x \right]_{0}^{\pi /4}\Rightarrow I=\frac{\pi }{4}\log 2\]You need to login to perform this action.
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