Definite Integration Question 157
Question: $ \int_0^{\pi }{x\log \sin x}dx= $
Options:
A) $ \frac{\pi }{2}\log \frac{1}{2} $
B) $ \frac{{{\pi }^{2}}}{2}\log \frac{1}{2} $
C) $ \pi \log \frac{1}{2} $
D) $ {{\pi }^{2}}\log \frac{1}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_0^{\pi }{x\log \sin xdx} $ ……(i) = $ \int_0^{\pi }{(\pi -x)\log \sin (\pi -x)dx} $ …..(ii)
By adding (i) and (ii), we get
$ 2I=\int_0^{\pi }{\pi }\log \sin xdx\Rightarrow I=\frac{2\pi }{2}\int_0^{\pi /2}{\log \sin xdx} $
$ =\pi ( \frac{\pi }{2}\log \frac{1}{2} )=\frac{{{\pi }^{2}}}{2}\log \frac{1}{2} $ .
BETA
