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\int _{{1}}^{{e^{{\pi }}}}{\frac  {\sin(\ln x)}{x}}\,dx\,

u=\ln x\,

du={\frac  {1}{x}}dx\,

So we have

\int _{{1}}^{{e^{{\pi }}}}{\frac  {\sin(\ln x)}{x}}\,dx=\int _{0}^{{\pi }}\sin u\,du\,

Notice, the limits of integration changed because when x=1\, and x=e^{{\pi }}\,, we have u=0\, and u=\pi \, respectively.

\int _{0}^{{\pi }}\sin u\,du=-\cos u{\bigg |}_{0}^{{\pi }}=-\cos \pi +\cos 0=2\,

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