Definite Integration Question 301
Question: $ \int_0^{\pi /4}{\log (1+\tan \theta )d\theta =} $
[SCRA 1986; Karnataka CET 2000, 05]
Options:
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} $
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Answer:
Correct Answer: C
Solution:
$ I=\int_0^{\pi /4}{\log (1+\tan \theta )d\theta } $
therefore $ I=\int_0^{\pi /4}{\log { 1+\tan ( \frac{\pi }{4}-\theta ) }}d\theta $
therefore I = $ \int_0^{\pi /4}{\log ( 1+\frac{1-\tan \theta }{1+\tan \theta } )d\theta } $
therefore 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} $ .
BETA
