_ 1 2 ^2 1 x , ( x - 1 x ) , , ,dx has the value equal to - Vidyadip

Question

$\int\limits_{\frac{1}{2}}^2 {\frac{1}{x}\,\sin \left( {x - \frac{1}{x}} \right)\,\,\,dx} $ has the value equal to

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Answer

a
$x = \frac{1}{t}$ $\Rightarrow dx = \,\frac{1}{{{t^2}}}\,dt$.
$I = \int\limits_2^{\frac{1}{2}} {t\,\,\sin \left( {\frac{1}{t} - t} \right)\,\left( { - \,\frac{1}{{{t^2}}}} \right)\,\,dt} $ $= \int\limits_2^{\frac{1}{2}} {\frac{1}{t}\,\sin \left( {t - \frac{1}{t}} \right)\,\,\,dt} $ $= -\int\limits_{\frac{1}{2}}^2 {\frac{1}{t}\,\sin \left( {t - \frac{1}{t}} \right)\,\,\,dt} $ $=$ $- I$
$\Rightarrow 2I = 0 \Rightarrow I = 0$
Alternatively : put $x = e^t \Rightarrow I =$ $\int\limits_{ - \ln 2}^{\ln 2} {\sin ({e^t} - {e^{ - t}})dt} $ $= 0$ $(odd\, function)$

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