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Relationship integration math problem

  1. Jul 9, 2014 #1
    This is about attempting to solve ##\left( y'\right)^2 = y^2 - 1 ##.

    [tex]\int\frac{dy}{\sqrt{y^2 -1}} = \pm \int dx[/tex]

    using a trig. substitution and another trick,

    [tex]\int\frac{dy}{\sqrt{y^2 -1}} = \ln\left(y \pm \sqrt{y^2 - 1} \right) + C[/tex]

    I'm not sure about that [itex]\pm[/itex] sign. It came in when doing [itex]\tan (\sec^{-1} y)[/itex] since I had [tex]\sec \theta = y \longrightarrow\tan^2 \theta = y^2 -1\longrightarrow \tan \theta = \pm \sqrt{y^2 -1}[/tex]
    The solution is supposed to be something like [itex]y = \cosh x[/itex]. But I am not sure how I'm supposed to be getting that from separating and integrating.

    [itex]\ln\left(y + \sqrt{y^2 - 1} \right) = \pm x + C [/itex]
    [itex]y + \sqrt{y^2 - 1} = C_1e^{\pm x} [/itex]
  2. jcsd
  3. Jul 9, 2014 #2
    Hello Misterx

    Please see the attached picture. Not sure if I am correct. I suggest waiting for someone to verify things :smile:

    Attached Files:

  4. Jul 9, 2014 #3


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    Homework Helper

    You can do [tex]
    \int \frac{1}{\sqrt{y^2 - 1}}\,dy
    [/tex] by substituting [itex]y = \cosh u[/itex] so that [itex]dy = \sinh u\,du[/itex] and using the relationship [tex]\cosh^2 u - \sinh^2 u = 1.[/tex]
    Alternatively, if you solve [tex]
    y = \cosh x = \frac{e^{x} + e^{-x}}{2}
    [/tex] for [itex]x = \mathrm{arccosh}(y) \geq 0[/itex] you will find that [tex]\mathrm{arccosh}(y) = \ln\left(y + \sqrt{y^2 - 1}\right).[/tex]

    Note that substituting [itex]y = \cosh u[/itex] gets you to the result with considerably less effort than substituting [itex]y = \sec \theta[/itex].
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