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Thread: Integration of inverse hyperbolic functions

  1. #1
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    Integration of inverse hyperbolic functions

    Prove that:
    \displaystyle \int^1_{\frac{4}{5}} arsech xdx=2\arctan 2-\frac{\pi}{2}-\frac{4}{5}\ln 2

    I found arsech x = \ln{\frac{1+\sqrt{1+x^2}}{x}}
    But When I tried to integrate it, it got very messy as well as not being the answer. I've got no idea how else to integrate, so any help will be very much appreciated.
    Thanks!
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  2. #2
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    Use integration by parts with u = \textrm{arsech}\,{x} and dv = 1.
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  3. #3
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    u=arsech x \rightarrow \frac{du}{dx}=-\frac{1}{x\sqrt{1-x^2}}
    \frac{dv}{dx}=1 \rightarrow v=x

    Then the integral
    x arsech x -\int -\frac{1}{\sqrt{1-x^2}}
    =x arsech x + \arcsin x

    Filling in the values, I get:
    \arctan \frac{4}{3}-\frac{\pi}{2}-\frac{4}{5}\ln 2
    Which is correct except for
    \arctan \frac{4}{3}
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