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