int_{log frac{1}{2}}^{log 2} sin left(frac{e^ | vedclass

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(D) Let $I = \int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^{x}-1}{e^{x}+1}\right) dx$.Note that $\log \frac{1}{2} = \log 1 - \log 2 = -\log 2$.S

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$0$

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(D) Let $I = \int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^{x}-1}{e^{x}+1}\right) dx$.Note that $\log \frac{1}{2} = \log 1 - \log 2 = -\log 2$.So,$I = \int_{-\log 2}^{\log 2} f(x) dx$,where $f(x) = \sin \left(\frac{e^{x}-1}{e^{x}+1}\right)$.Check if $f(x)$ is an odd function: $f(-x) = \sin \left(\frac{e^{-x}-1}{e^{-x}+1}\right) = \sin \left(\frac{\frac{1}{e^{x}}-1}{\frac{1}{e^{x}}+1}\right) = \sin \left(\frac{1-e^{x}}{1+e^{x}}\right) = \sin \left(-\frac{e^{x}-1}{e^{x}+1}\right) = -\sin \left(\frac{e^{x}-1}{e^{x}+1}\right) = -f(x)$.Since $f(x)$ is an odd function and the interval is symmetric $[-\log 2, \log 2]$,the integral evaluates to $0$.

$\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^{x}-1}{e^{x}+1}\right) dx=$

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