Mathematics

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

$0$

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.