Mathematics

# The value of  $\int { \sqrt { \dfrac { { e }^{ x }-1 }{ { e }^{ x }+1 } } } dx$ is equal to

##### SOLUTION
$I=\int { \sqrt { \dfrac { { e }^{ x }-1 }{ { e }^{ x }+1 } } } dx$
Put $e^x=t$
$\therefore { e }^{ x }dx=dt\\ dx=\dfrac { dt }{ { e }^{ x } } =\dfrac { dt }{ t } \\ \therefore I=\int { \dfrac { 1 }{ t } \sqrt { \dfrac { t-1 }{ t+1 } } } dt$
Put $t=\sec \theta$
$\therefore dt=\sec { \theta } \tan { \theta } d\theta \\ I=\int { \dfrac { 1 }{ \sec { \theta } } \sqrt { \dfrac { \sec { \theta } -1 }{ \sec { \theta } +1 } } } \times \sec { \theta } \tan { \theta } d\theta$
$I=\int { \dfrac { \sec { \theta } -1 }{ \sqrt { \sec ^{ 2 }{ \theta } -1 } } } \tan { \theta } d\theta$ (muliplying neumerator and denomenator by $\sec\theta-1$)
$I=\int { \dfrac { \left( \sec { \theta } -1 \right) }{ \tan { \theta } } } \tan { \theta } d\theta \quad (\therefore \sec ^{ 2 }{ \theta } -1=\tan ^{ 2 }{ \theta } )\\ I=\int { \left( \sec { \theta } -1 \right) } d\theta \\ I=\ln { \left| \sec { \theta } +\tan { \theta } \right| } -\theta +c\quad (\because \int { \left( \sec { \theta } \right) } d\theta =\left| n \right| \sec { \theta } +\tan { \theta } )\\ I=\ln { \left| { e }^{ x }+\sqrt { { e }^{ 2x }-1 } \right| } -\sec ^{ -1 }{ { (e }^{ x }) } +c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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