Question

${\int }_{\log 1/2}^{\log 2}\sin \left(\frac{e^x-1}{e^x+1}\right)dx$ is equal to

• A. $\cos \frac{1}{3}$
• B. $0$
• C. $2\cos 2$
• D. none of these

Answer 1

Answer: (B)

$0$

Let $I={\int }_{\log 1/2}^{\log 2}\sin \left(\frac{e^x-1}{e^x+1}\right)dx$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_{\log 1/2}^{\log 2}\sin \left(\frac{e^{-x}-1}{e^{-x}+1}\right)dx=-{\int }_{\log 1/2}^{\log 2}\sin \left(\frac{e^x-1}{e^x+1}\right)dx=-I\therefore 2I=0\Rightarrow I=0$