Explanation
$$\displaystyle{I=\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx.....(i)}$$
Let $$t=\dfrac { 1 }{ x } $$
$$dt=-\dfrac { 1 }{ { x }^{ 2 } } dx\\ dx={ -x }^{ 2 }dt$$
Substituting in $$(i)$$
$$\displaystyle{I=\int _{ \pi }^{ \frac { 1 }{ \pi } }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } (-x^{ 2 })dt\\ \\ I=-\int _{ \pi }^{ \frac { 1 }{ \pi } }{ x } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dt\\ I=-\int _{ \pi }^{ \frac { 1 }{ \pi } }{ \dfrac { 1 }{ t } } .\sin { \left( \dfrac { 1 }{ t } -t \right) } dt\\ I=-\int _{ \pi }^{ \frac { 1 }{ \pi } }{ \dfrac { 1 }{ t } } .\sin { \left( -(t-\dfrac { 1 }{ t } ) \right) } dt}$$
using $$\sin { (-x)=-\sin { x } } $$
$$\displaystyle{I=\int _{ \pi }^{ \frac { 1 }{ \pi } }{ \dfrac { 1 }{ t } } .\sin { \left( t-\dfrac { 1 }{ t } \right) } dt}$$
Using $$\displaystyle{\int _{ a }^{ b }{ f\left( x \right) dx=-\int _{ b }^{ a }{ f\left( x \right) dx } } }$$ and changing the variable from $$t$$ to $$x$$
$$\displaystyle{I=-\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx.....(ii)}$$
Adding $$(i)$$ and $$(ii)$$
$$\displaystyle{2I=\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx-\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx=0\\ \Rightarrow I=0}$$
So option $$(A)$$ is correct.
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