Definite Integration Question 92
Question: If $ \int_0^{\pi }{xf(\sin x)dx=A}\int_0^{\pi /2}{f(\sin x)dx} $ , then A is
[AIEEE 2004]
Options:
A) $ 2\pi $
B) $ \pi $
C) $ \frac{\pi }{4} $
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ I=\int_0^{\pi }{xf(\sin x)dx}=A\int_0^{\pi /2}{f(\sin x)dx} $
Now, $ 2I=\int_0^{\pi }{xf(\sin x)dx+\int_0^{\pi }{(\pi -x)f[\sin (\pi -x)]dx}} $
$ =\int_0^{\pi }{\pi f(\sin x)dx}=\pi \int_0^{\pi }{f(\sin x)dx} $
therefore $ 2I=2\pi \int_0^{\pi /2}{f(\sin x)dx} $
$ I=\pi \int_0^{\pi /2}{f(\sin x)dx} $
$ =A\int_0^{\pi }{f(\sin x)dx} $ .
Hence $ A=\pi $ .
BETA
