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Impossible Integral?

  1. Dec 10, 2011 #1
    This is NOT homework, I am not in this crazy of a class...


    1. The problem statement, all variables and given/known data
    [itex]\displaystyle \int^{45}_{0}\frac{\sin^2(\arcsin(\frac{t \sin \gamma }{ r })-\gamma+180)-\sin^2(\arcsin(\frac{t\sin\gamma}{r})-\gamma)}{\sin^2\gamma}\Delta\gamma[/itex]

    [itex]r = 1, t = \sqrt{2}[/itex]

    2. Relevant equations
    The above.


    3. The attempt at a solution
    I cannot even come close to attempting this, but I believe the answer should be 360... If anyone is able to do this I would love to see it worked out and to know if 360 is correct.
     
  2. jcsd
  3. Dec 10, 2011 #2

    LCKurtz

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    Not knowing where this problem came from, I would almost bet that it should be phrased in radians instead of, apparently, degrees. If so, then if you call[tex]
    \alpha = \arcsin(\frac{t \sin \gamma )}{ r }-\gamma [/tex]then the numerator is[tex]
    \sin^2(\alpha + \pi)-\sin^2(\alpha) = 0[/tex]giving 0 for the answer to the integral.
     
  4. Dec 10, 2011 #3
    Well....thats no good! I am not sure when to use radians versus degrees. It shouldn't make a difference should it, other than perhaps one being easier than the other?
     
  5. Dec 11, 2011 #4
    There's nothing wrong with how you've used degrees instead of radians in your integral. What matters is that you're consistent and the arguments of each have the right meaning. Since you are varying gamma in degrees, there's no issue with it. LCKurtz comment is the exactly same when replacing pi with 180.

    There is potentially one issue that could arise, though, in the method above, though I can't say for certain at the moment whether or not it comes into play. This is the fact that sine is 0 when gamma is 0, so you get a situation where your integrand is 0/0. You'll want to check the limit as gamma goes to 0 in the integrand and make sure it's not infinite. In either case, unless you show that the integrand goes to 0 there in the limit and "fill in" the discontinuity there (much like you can with sin(x)/x), it might be better to represent the integral as from [itex]0^+[/itex] to 45.
     
  6. Dec 11, 2011 #5
    That wouldn't save it from being 0 though right? it would just be a more correct 0?
     
  7. Dec 11, 2011 #6
    I'd start off with
    Sin[x+180]=-sin[x]
    so you'd have the integrand as

    [itex]\frac{-2sin^2(arcsin(\frac{t sin(\gamma)}{r})-\gamma )}{sin^2(\gamma )}[/itex]
    and then use the sin double angle formula to get
    [itex]\alpha = arcsin(stuff)[/itex]
    [itex]\beta = sinstuff[/itex]
    [itex]sin^2(\alpha-\gamma)=(sin(\alpha)cos(\gamma)-cos(\alpha)sin(\gamma))^2[/itex]
    [itex]=sin^2(\alpha)cos^2(\gamma)-2sin(\alpha)cos(\alpha)sin(\gamma)cos(\gamma)+cos^2(\alpha)sin^2(\gamma)[/itex]
    [itex]=\beta^2cos^2(\gamma)-2\beta cos(\alpha)sin(\gamma)cos(\gamma)+cos^2(\alpha)sin^2(\gamma)[/itex]
    and the use the sin=sqrt 1-cos
    [itex]=\beta^2cos^2(\gamma)-2\beta \sqrt{1-\beta^2}sin(\gamma)cos(\gamma)+(1-\beta^2)sin^2(\gamma)[/itex]

    which gets rid of all the arcsin stuff and it becomes a problem of integrating sums and products of sins and cosins which shouldn't be too hard

    [itex]-2\int_{0}^{45}\frac{\beta^2cos^2(\gamma)-2\beta \sqrt{1-\beta^2}sin(\gamma)cos(\gamma)+(1-\beta^2)sin^2(\gamma)}{sin^2(\gamma)} = -2\int_{0}^{45}\beta^2 cotan^2(\gamma) d\gamma
    +4\int_{0}^{45}\beta \sqrt{1-\beta^2} cotan(\gamma) d\gamma-2\int_{0}^{45}(1-\beta)^2d\gamma[/itex]

    which you should be able to do

    good luck!

    EDIT;
    nvm, I'm talking out of my *** since it was sin^2 rather than just sin so sin^2(x+180)-sin^2(x)=0 so the integral is zero for all x
     
    Last edited: Dec 11, 2011
  8. Dec 11, 2011 #7
    Using [itex]0^+[/itex] would simply make it so that the integrand is defined everywhere in your integration. It still would diverge to infinity if the limit as gamma went to 0 sent the integrand to infinity, though.
     
  9. Dec 11, 2011 #8

    LCKurtz

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    But there could easily be an issue with it. The reason for using radians in such an integral is that if by chance you should be able to solve it explicitly with antiderivative techniques, you wouldn't be able to use the standard formulas for the trig functions. If you don't know what you are doing you will get incorrect answers to even simple integrals using degrees. The antiderivative of cos(x) is not sin(x) if x is measured in degrees.
     
  10. Dec 11, 2011 #9
    Interesting...so at any rate it would maybe be a good habit to be in, using radians?
     
  11. Dec 11, 2011 #10

    LCKurtz

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    No "maybe" about it.
     
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