How to integrate $$\int_{1/2}^{2}\frac{\sin(\frac{x^{2}-1}{x})}{x} dx$$ I tried a couple of thing such as

  • integration by parts by letting $\frac{1}{x}$ as an intigrable function but got $$-\int_{1/2}^{2}\cos(x-\frac{1}{x})*(1+\frac{1}{x^2})*\ln{x}$$ I don't know how to proceed further please help
  • 4
    $\begingroup$ If you substitute $u=\frac{1}{x}$, you get the same integral, but with a minus sign. So the integral is $0$ $\endgroup$ – Jakobian Nov 11 '18 at 15:29

$$I=\int_{\frac12}^{2}\frac{\color{blue}{\sin\left(x-\frac1{x}\right)}}{x} dx$$ $$\text{let}\ x=\frac{1}{t}\rightarrow dx=-\frac{1}{t^2}dt$$ $$I=\int^{\frac12}_2 \frac{\sin\left(\frac1{t}-t\right)}{\frac{1}{t}}\frac{-dt}{t^2}=\int_{\frac12}^2\frac{\sin\left(\frac1{t}-t\right)}{t}dt\overset{t=x}=-\int_{\frac12}^2\frac{\color{red}{\sin\left(x-\frac1{x}\right)}}{x}dx$$ Thus, if we add with the initial integral we get: $$2I=\int_{\frac12}^2\frac{\overset{=0}{\color{blue}{\sin\left(x-\frac1{x}\right)}-\color{red}{\sin\left(x-\frac1{x}\right)}}}{x}dx=0\Rightarrow I=0$$


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