<s>[INST] How to evaluate this integral $\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}$? 
Evaluate
  $$\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}.$$

I am unable to think of any way to solve this.
 [/INST] $$I=\int_{\frac{1}{3}}^3 \dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx=\underbrace{\int_{\frac{1}{3}}^1\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx}_J+\underbrace{\int_1^3\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx}_H$$
$$J=\int_{\frac{1}{3}}^1\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx\quad t=\dfrac{1}{x}\rightarrow x=\dfrac{1}{t}\rightarrow dx=-\dfrac{1}{t^2}\quad \text{and}\quad \left\{ \begin{array}{lcc}
             x=\dfrac{1}{3}\rightarrow t=3 \\
             \\ x=1\rightarrow t=1 
             \end{array}
   \right.$$
$$J=\int_3^1\dfrac{\sin \left( t-\dfrac{1}{t}\right)}{\dfrac{1}{t}}\left( -\dfrac{1}{t^2}\right)dt=-\int_1^3\dfrac{\sin \left[ -\left( \dfrac{1}{t}-t\right) \right]}{-t}dt=-\int_1^3 \dfrac{-\sin \left( \dfrac{1}{t}-t\right)}{-t}dt=$$
$$=-\int_1^3\dfrac{\sin \left( \dfrac{1}{t}-t\right)}{t}dt=-\int_1^3\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx=-H\rightarrow I=-H+H=\boxed{0}$$
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