Mathematics

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


ANSWER

$$0$$


SOLUTION
Let $$\displaystyle I=\displaystyle \int _{ \log  1/2 }^{ \log  2 } \sin  \left( \frac { e^{ x }-1 }{ e^{ x }+1 }  \right) dx$$
Using $$\displaystyle \int _{ a }^{ b }{ f\left( x \right) dx } =\displaystyle \int _{ a }^{ b }{ f\left( a+b-x \right) dx } $$
$$\displaystyle I=\displaystyle \int _{ \log  1/2 }^{ \log  2 } \sin  \left( \frac { e^{ -x }-1 }{ e^{ -x }+1 }  \right) dx=-\displaystyle \int _{ \log  1/2 }^{ \log  2 } \sin  \left( \frac { e^{ x }-1 }{ e^{ x }+1 }  \right) dx=-I\\ \therefore 2I=0\Rightarrow I=0$$
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Single Correct Medium Published on 17th 09, 2020
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