Mathematics

# If $\displaystyle \int_{0}^{1}\frac{dt}{e^{t}+e^{-t}}=\tan^{-1}p$ then $p$, equals

##### ANSWER

$\displaystyle \frac{e-1}{e+1}$

##### SOLUTION
$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dt }{ e^{ t }+e^{ -t } } =\int _{ 0 }^{ 1 } \frac { e^{ t }dt }{ 1+e^{ 2t } }$
Substitute ${ e }^{ t }=u\Rightarrow { e }^{ t }dt=du$
$\displaystyle \therefore I=\int _{ 1 }^{ e }{ \frac { 1 }{ 1+{ u }^{ 2 } } du } ={ \left[ \tan ^{ -1 }{ u } \right] }_{ 1 }^{ e }$
$\displaystyle =\tan ^{ -1 }{ e } -\tan ^{ -1 }{ 1 } =\tan ^{ -1 }{ \left( \frac { e-1 }{ 1+e } \right) }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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