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

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(A) Let $I = \int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x-1}{e^x+1}\right) d x$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+

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(A) Let $I = \int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x-1}{e^x+1}ight) d x$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we note that the limits are $a = \log \frac{1}{2} = -\log 2$ and $b = \log 2$. Thus,$a+b = 0$. So,$I = \int_{-\log 2}^{\log 2} \sin \left(\frac{e^{-x}-1}{e^{-x}+1}ight) d x$. Simplifying the argument of the sine function: $\frac{e^{-x}-1}{e^{-x}+1} = \frac{\frac{1}{e^x}-1}{\frac{1}{e^x}+1} = \frac{1-e^x}{1+e^x} = -\left(\frac{e^x-1}{e^x+1}ight)$. Since $\sin(- 	heta) = -\sin(	heta)$,the integrand is an odd function. Therefore,$I = \int_{-\log 2}^{\log 2} -\sin \left(\frac{e^x-1}{e^x+1}ight) d x = -I$. This implies $2I = 0$,so $I = 0$.

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

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