Definite Integration Question 179
Question: $ \int_0^{\pi /2}{{}}\log \sin xdx= $
[MP PET 1994; RPET 1995, 96, 97]
Options:
A) $ -( \frac{\pi }{2} )\log 2 $
B) $ \pi \log \frac{1}{2} $
C) $ -\pi \log \frac{1}{2} $
D) $ \frac{\pi }{2}\log 2 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_0^{\pi /2}{\log \sin xdx=\int_0^{\pi /2}{\log \cos xdx}} $
therefore $ 2I=\int_0^{\pi /2}{\log \sin x\cos xdx}=\int_0^{\pi /2}{\log \sin 2xdx}-\int_0^{\pi /2}{\log 2dx} $
$ =\frac{1}{2}\int_0^{\pi }{\log \sin tdt-\frac{\pi }{2}\log 2} $ , (Putting $ 2x=t $ ) $ =\frac{1}{2}.2\int_0^{\pi /2}{\log \sin tdt-\frac{\pi }{2}\log 2} $
$ \Rightarrow 2I=I-\frac{\pi }{2}\log 2\Rightarrow I=\frac{-\pi }{2}\log 2 $ , $ { \because \int_a^{b}{f(x)dx=\int_a^{b}{f(t)dt}} } $ .
BETA
