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    • Thread Starter
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    Trying to differentiate:

    1/cosh(x)

    I know what it should end up as:

    -sinh(x) / cosh^2(x)

    but I don't know how to get to it.

    Thanks.
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    (Original post by mackemforever)
    Trying to integrate:

    1/cosh(x)

    I know what it should end up as:

    -sinh(x) / cosh^2(x)

    but I don't know how to get to it.

    Thanks.
    That's the derivative of 1/cosh(x) and not the integral.
    To evaluate this integral rewrite it as:

    I = 2\displaystyle\int \dfrac{1}{e^x+e^{-x}}dx = 2\displaystyle\int \dfrac{e^x}{e^{2x}+1}dx.

    Then use a substitution of u=e^{x} if you can't spot by inspection that this integral leads to arctan.
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    \int \frac{1}{\cosh{x}} dx = 2 \int \frac{1}{e^x+e^{-x}} dx

    let u=e^x= \frac{du}{dx}

    you should then be able to rearrange and get it to

    \int \frac{du}{1+u^2}

    which is a standard integral
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    (Original post by mackemforever)
    Trying to integrate:

    1/cosh(x)

    I know what it should end up as:

    -sinh(x) / cosh^2(x)

    but I don't know how to get to it.

    Thanks.
    To solve \displaystyle \int R(sinhx, coshx)\ dx
    where R is a fuction formed with rational operations from
    the given functions and constans, use the substittion of
    \displaystyle t=tanh\frac{x}{2}
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    Crap, sorry, I meant differentiate!

    My bad.
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    (Original post by mackemforever)
    Crap, sorry, I meant differentiate!

    My bad.
    In which case, you should use the chain rule.
    i.e. note that:
    \frac{d}{dx}[\frac{1}{\cosh x}] = \frac{d}{dx}[(\cosh x)^{-1}].
    Now if you let u=\cosh x then \frac{du}{dx}=\sinh x and our original derivative becomes \frac{d}{du}[u^{-1}]. Then note that \frac{d}{dx}[\frac{1}{\cosh x}] = \frac{du}{dx} \times \frac{d}{du}[u^{-1}]
 
 
 
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