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Integrating sine^x

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve by variation of parameters:
    y" + 3y' + 2y = sinex

    2. Relevant equations
    Finding the complimentary yields:
    yc = c1e-x + c2e-2x

    3. The attempt at a solution
    I set up the Wronskians and got:
    μ1 = ∫e-2xsin(ex)dx
    μ2 = -∫e-xsin(ex)dx

    The problem is that I have no idea how to integrate sin(ex).

    I tried subbing u = e-x du = -e-x for μ1
    => ∫-u du sin(u-1)
    Integration by parts either attempts to integrate sin(u-1) or endlessly integrates u without repeat.

    That failed, so I tried just integrating by parts of μ1; it took 2 repetitions to get:
    μ1 = -½e-2xsin(ex) - ½e-xcos(ex) + ∫½sin(ex)dx
    I thought it might work if I get -ex∫½sin(ex)] in μ2
    μ2 = e-xsin(ex) - ∫cos(ex)dx

    Going further into the integration by parts just adds more complications, such as adding "x" as a term as well as going into higher powers of ex.

    I can't express the integral as a series; that's next chapter and not covered on the mid-term in a few days (I'm currently hoping the mid-term doesn't have this problem).
     
  2. jcsd
  3. Oct 21, 2014 #2

    Simon Bridge

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    Did you try substitution: ##u=e^x##.
     
  4. Oct 21, 2014 #3
    I did initially, but found it problematic since it ends up becoming ½∫sinu/u. If you keep on doing it it winds up being a giant beast of an equation which doesn't appear to match μ2 at all.

    I thought part of it might equal μ2, but that didn't happen either. The two just keep mismatching cosine and sine functions endlessly.

    I just realized I did the Wronskians wrong; the bottom isn't 1... And it ends up a LOT easier, but I still can't quite seem to get it right.
    With the correct Wronskian I instead got ∫sin(u)/u for μ1 and a nice and short -½cos(ex) for μ2. This does cause an elimination between the two equations, but I'm still getting a vast number of functions for μ1 that doesn't seem to have an end.

    I'm going to check back on this in the morning.
     
  5. Oct 21, 2014 #4

    Ray Vickson

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    The integral ##\int \sin(e^x) \, dx = \text{Si}(e^x) + C##, where ##\text{Si}## is a "non-elementary" function.
     
  6. Oct 21, 2014 #5
    Resolved: I kept working with substituting ex = u on both μ's and it actually works insanely well.
     
  7. Oct 21, 2014 #6

    Simon Bridge

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    Well done - getting the correct Wronskian is helpful too of course ;)
    (If the Wronskian was incorrect, does that make it a Wrightskian?)

    Aside:
    The "non-elementary function" Si(x) in post #4 is called the Sine Integral and you can look it up.

    For that matter, sin(x)/x is called "sinc(x)" ... and you can look that up too.
     
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