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

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Aicharya · Accepted Answer

$\begin{aligned} & \int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{e}^{\mathrm{x}}+1}ight) \mathrm{dx}=\int_{-\log 2}^{\log 2} \sin \left(\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{e}^{\mathrm{x}}+1}ight) \mathrm{dx}=0 \ & {\left[\because \sin \left(\frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{e}^{\mathrm{x}}+1}ight) 	ext { is an odd function }ight]}\end{aligned}$

$\int_{\log \frac{1}{2}}^{\log 2} \sin…

Solution