Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_{1/\pi }^{2/\pi } {\frac{{\sin \left( {1/x} \right)}}{{{x^2}}}} dx \cr
& {\text{Let }}u = \frac{1}{x},{\text{ }}du = - \frac{1}{{{x^2}}}dx \cr
& {\text{The new limits of integration are:}} \cr
& x = \frac{2}{\pi } \to u = \frac{1}{{2/\pi }} = \frac{\pi }{2} \cr
& x = \frac{1}{\pi } \to u = \frac{1}{{1/\pi }} = \pi \cr
& {\text{Substituting}} \cr
& \int_{1/\pi }^{2/\pi } {\frac{{\sin \left( {1/x} \right)}}{{{x^2}}}} dx = \int_\pi ^{\pi /2} {\sin u\left( { - du} \right)} \cr
& = - \int_\pi ^{\pi /2} {\sin udu} \cr
& {\text{Integrating}} \cr
& = \left[ {\cos u} \right]_\pi ^{\pi /2} \cr
& = \cos \left( {\frac{\pi }{2}} \right) - \cos \left( \pi \right) \cr
& = 1 \cr} $$
