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Homework Help: Hyperbolic Substituion. I am wrong?

  1. Aug 7, 2011 #1
    1. The problem statement, all variables and given/known data


    [tex]\int \;\sinh(2x) \cosh(2x) dx[/tex]



    3. The attempt at a solution

    I let u = sinh(2x), du = 2cosh(2x)dx

    Integrating I shuold get


    [tex]\frac{1}{4} sinh^2 (2x) + C[/tex]

    But http://www.wolframalpha.com/input/?i=Integrate[cosh%282x%29%28sinh%282x%29%29%2Cx]

    says I need to let u = 2x first. Why? I don't think i am wrong, my algebra looked correct.
     
  2. jcsd
  3. Aug 7, 2011 #2

    HallsofIvy

    User Avatar
    Science Advisor

    No, you don't need to do that, you just can.

    If you let u= 2x, then, of course, dx= (1/2) du so your integral becomes
    [tex]\frac{1}{2}\int sinh(u)cosh(u)du[/tex]
    Then let v= sinh(u) so that dv= cosh(u)du and the integral becomes
    [tex]\frac{1}{2}\int v dv= \frac{1}{4}v^2+ C= \frac{1}{4}sinh^2(u)+ C= \frac{1}{4}sinh^2(2x)+ C[/tex]
    just as you have.
    That's just using two consecutive substitutions rather than combining them as you did.

    Of course, you could do it the other way two: let u= cosh(2x) so that du= 2sinh(x)dx and the integral becomes
    [tex]\frac{1}{2}u^2du= \frac{1}{4}cosh^2(2x)+ D[/tex]

    Because [itex]cosh^2(2x)= sinh^2(2x)+ 1[/itex] that is exactly the same thing with a difference constant.

    I believe your wolframalpha reference gives the integral as
    [tex]\frac{1}{8}cosh(4x)+ Constant[/tex]
    which looks very different but is actually the same as your result.

    Remember that
    [tex]sinh(x)= \frac{e^x- e^{-x}}{2}[/tex]

    Squaring that:
    [tex]sinh^2(x)= \frac{(e^x)^2- 2e^xe^{-x}+ (e^{-x})^2}{4}= \frac{e^{2x}- 2+ e^{-2x}}{4}[/tex]
    [tex]= \frac{1}{2}\frac{e^{2x}+ e^{-2x}}{2}- \frac{1}{2}= \frac{1}{2}cosh(2x)- \frac{1}{2}[/tex]
    so that
    [tex]\frac{1}{4}sinh^2(2x)= \frac{1}{8}cosh(4x)- \frac{1}{8}[/tex]
    differing from the original form only by a constant.
     
    Last edited by a moderator: Aug 7, 2011
  4. Aug 7, 2011 #3
    Oh right, we are doing an indefinite integral here.
     
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