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
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