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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