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Math Help - Integrate inverse of cosh

  1. #1
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    Integrate inverse of cosh

    I have an equation i wish to integrate

    <br />
\int\frac{dx}{coshx+1}<br />

    What is the best way to start a problem like this. Should i substitute or integrate by parts? Any help appreciated!!

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  2. #2
    Eater of Worlds
    galactus's Avatar
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    Try using its e identity.

    cosh(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)

    \frac{1}{\frac{1}{2}\left(e^{x}+e^{-x}\right)+1}=

    \frac{2e^{x}}{e^{2x}+2e^{x}+1}=\frac{2}{e^{x}+1}-\frac{2}{(e^{x}+1)^{2}}

    2\int\frac{1}{e^{x}+1}dx-2\int\frac{1}{(e^{x}+1)^{2}}dx
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    but \frac{2e^{x}}{e^{2x}+2e^{x}+1}=\frac{2e^{x}}{\left  ( e^{x}+1 \right)^{2}}.
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  4. #4
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    I don't understand how you got to

    Quote Originally Posted by galactus View Post

    \frac{2e^{x}}{e^{2x}+2e^{x}+1}=\frac{2}{e^{x}+1}-\frac{2}{(e^{x}+1)^{2}}
    from this

    Quote Originally Posted by galactus View Post

    \frac{1}{\frac{1}{2}\left(e^{x}+e^{-x}\right)+1}
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  5. #5
    Eater of Worlds
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    It's just a little algebra.

    \frac{2u}{(u+1)^{2}}

    \frac{2u+2-2}{(u+1)^{2}}

    \frac{2(u+1)-2}{(u+1)^{2}}

    \frac{2}{u+1}-\frac{2}{(u+1)^{2}}

    Now, plug the e back in.
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