Explanation
I=∫π1π1x.sin(x−1x)dx.....(i)
Let t=1x
dt=−1x2dxdx=−x2dt
Substituting in (i)
I=∫1ππ1x.sin(x−1x)(−x2)dtI=−∫1ππx.sin(x−1x)dtI=−∫1ππ1t.sin(1t−t)dtI=−∫1ππ1t.sin(−(t−1t))dt
using sin(−x)=−sinx
I=∫1ππ1t.sin(t−1t)dt
Using ∫baf(x)dx=−∫abf(x)dx and changing the variable from t to x
I=−∫π1π1x.sin(x−1x)dx.....(ii)
Adding (i) and (ii)
2I=∫π1π1x.sin(x−1x)dx−∫π1π1x.sin(x−1x)dx=0⇒I=0
So option (A) is correct.
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