Definite Integration Question 356
Question: The value of integral $ \int _{1/\pi }^{2/\pi }{\frac{\sin (1/x)}{x^{2}}}dx= $
[IIT 1990]
Options:
A) 2
B) $ -1 $
C) 0
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
Put $ t=\frac{1}{x}\Rightarrow dt=-\frac{1}{x^{2}}dx $ as $ t=\frac{\pi }{2} $ and $ \pi $
$ \therefore $ $ \int _{1/\pi }^{2/\pi }{\frac{\sin ( \frac{1}{x} )}{x^{2}}dx} $
$ =-\int _{\pi /2}^{\pi }{\sin tdt=-[\cos t] _{\pi /2}^{\pi }} $
$ =-[ \cos \pi -\cos ( \frac{\pi }{2} ) ]=1 $ .
BETA
