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Q.
$\int_{\log\frac{1} {2}} ^{\log2} \, \sin (\frac {e^x-1} {e^x+1})dx$ is equal to
Integrals
Solution:
Let $f \left(x\right)=sin \left(\frac{e^{x}-1}{e^{x}+1}\right)$
$\therefore f\left(-x\right)=sin\left(\frac{e^{-x}-1}{e^{-x}+1}\right)=sin \left(\frac{\frac{1}{e^{x}}-1}{\frac{1}{e^{x}}+1}\right)$
$=sin \left(\frac{1-e^{x}}{1+e^{x}}\right)=-sin \left(\frac{e^{x}-1}{e^{x}+1}\right)=-f \left(x\right)$
$\therefore f \left(x\right)$ is an odd function.
$\therefore $ given integral $=\int_{-log\,2}^{log\,2} sin \left(\frac{e^{x}-1}{e^{x}+1}\right)dx=0$