Definite Integration Question 168
Question: $ \int_0^{\pi /2}{\log \tan xdx=} $
[MP PET 1999; RPET 2001, 02; Karnataka CET 1999, 2000, 01, 02]
Options:
A) $ \frac{\pi }{2}{\log _{e}}2 $
B) $ -\frac{\pi }{2}{\log _{e}}2 $
C) $ \pi {\log _{e}}2 $
D) 0
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Answer:
Correct Answer: D
Solution:
$ \int_0^{\pi /2}{\log \tan xdx=}\int_0^{\pi /2}{\log ( \frac{\sin x}{\cos x} )dx} $
$ =\int_0^{\pi /2}{\log \sin xdx-\int_0^{\pi /2}{\log \cos xdx=0}} $ , $ { \because \int_0^{a}{f(x)dx=\int_0^{a}{f(a-x)dx}} } $ .
BETA
