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$\int_{\frac{1}{3}}^3 \frac{1}{x} \sin \left(\frac{1}{x}-x\right) d x=$
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Let
$$
\begin{aligned}
& I=\int_{1 / 3}^3 \frac{1}{x} \sin \left(\frac{1}{x}-x\right) d x \\
& \text { Put } x=\frac{1}{t} \Rightarrow d x=\frac{-1}{t^2} d t \\
& x=\frac{1}{3}, t=3 \text { and } x=3, t=\frac{1}{3} \\
& \therefore I=\int_3^{1 / 3} t \sin \left(t-\frac{1}{t}\right)\left(\frac{-1}{t^2}\right) d t \\
& I=-\int_3^{1 / 3} \frac{1}{t} \sin \left(t-\frac{1}{t}\right) d t \\
& \Rightarrow \quad I=\int_3^{1 / 3} \frac{1}{t} \sin \left(\frac{1}{t}-t\right) d t \\
& I=-\int_{1 / 3}^3 \frac{1}{t} \sin \left(\frac{1}{t}-t\right) d t \\
& \Rightarrow \quad I=-I \\
& \Rightarrow \quad I=0 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& I=\int_{1 / 3}^3 \frac{1}{x} \sin \left(\frac{1}{x}-x\right) d x \\
& \text { Put } x=\frac{1}{t} \Rightarrow d x=\frac{-1}{t^2} d t \\
& x=\frac{1}{3}, t=3 \text { and } x=3, t=\frac{1}{3} \\
& \therefore I=\int_3^{1 / 3} t \sin \left(t-\frac{1}{t}\right)\left(\frac{-1}{t^2}\right) d t \\
& I=-\int_3^{1 / 3} \frac{1}{t} \sin \left(t-\frac{1}{t}\right) d t \\
& \Rightarrow \quad I=\int_3^{1 / 3} \frac{1}{t} \sin \left(\frac{1}{t}-t\right) d t \\
& I=-\int_{1 / 3}^3 \frac{1}{t} \sin \left(\frac{1}{t}-t\right) d t \\
& \Rightarrow \quad I=-I \\
& \Rightarrow \quad I=0 \\
&
\end{aligned}
$$
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