Definite Integration Question 48
Question: Let $ I_1=\int_a^{\pi -a}{xf(\sin x)dx,I_2=\int_a^{\pi -a}{f(\sin x)dx}} $ , then $ I_2 $ is equal to
[AMU 2000]
Options:
A) $ \frac{\pi }{2}I_1 $
B) $ \pi I_1 $
C) $ \frac{2}{\pi }I_1 $
D) $ 2I_1 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ I_1=\int_a^{\pi -a}{xf(\sin x)dx} $
$ =\int_a^{\pi -a}{(\pi -x)f(\sin (\pi -x))dx} $ , $ [\because \int_a^{b}{f(x)dx=\int_a^{b}{f(a+b-x)dx}}] $
$ =\int_a^{\pi -a}{(\pi -x)f(\sin x)dx} $
$ =\int_a^{\pi -a}{\pi f(\sin x)dx-I_1} $
$ \Rightarrow 2I_1=\pi I_2\Rightarrow I_2=\frac{2}{\pi }I_1 $ .
BETA
