A) 2
B) \[-1\]
C) 0
D) 1
Correct Answer: D
Solution :
Put \[t=\frac{1}{x}\Rightarrow dt=-\frac{1}{{{x}^{2}}}dx\]as \[t=\frac{\pi }{2}\]and \[\pi \] \[\therefore \] \[\int_{1/\pi }^{2/\pi }{\frac{\sin \left( \frac{1}{x} \right)}{{{x}^{2}}}dx}\]\[=-\int_{\pi /2}^{\pi }{\sin t\,dt=-[\cos t]_{\pi /2}^{\pi }}\] \[=-\left[ \cos \pi -\cos \left( \frac{\pi }{2} \right) \right]=1\].
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