Solveeit Logo
Login

Question

Mathematics Question on integral

∫log⁡12log⁡2 sin⁡(ex−1ex+1)dx\int_{\log\frac{1} {2}} ^{\log2} \, \sin (\frac {e^x-1} {e^x+1})dx is equal to

A

cos⁡13\cos \frac {1} {3}

B

0

C

2cos⁡22 \cos 2

D

none of these

Answer

0

Explanation

Solution

Let f(x)=sin(ex−1ex+1)f \left(x\right)=sin \left(\frac{e^{x}-1}{e^{x}+1}\right) ∴f(−x)=sin(e−x−1e−x+1)=sin(1ex−11ex+1)\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(1−ex1+ex)=−sin(ex−1ex+1)=−f(x)=sin \left(\frac{1-e^{x}}{1+e^{x}}\right)=-sin \left(\frac{e^{x}-1}{e^{x}+1}\right)=-f \left(x\right) ∴f(x)\therefore f \left(x\right) is an odd function. ∴\therefore given integral =∫−log 2log 2sin(ex−1ex+1)dx=0=\int_{-log\,2}^{log\,2} sin \left(\frac{e^{x}-1}{e^{x}+1}\right)dx=0