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Stuck on an Integral Problem

  1. Nov 25, 2009 #1
    1. The problem statement, all variables and given/known data
    I was messing around with various integral problems for a test a few months before, and I got stuck on a problem I made up:
    http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/e/f/2/ef2e9abca8945a10d5df45e7b73755ecb34cf21d.gif [Broken]


    2. Relevant equations
    No idea. I used everything in my Mathematics arsenal of Calculus BC.


    3. The attempt at a solution
    I'm very curious on learning on how to solve this. I developed some theorems and methods along the summer that ended up being already formed and used with slightly different notation. For example, I used a variant of U-substitution to solve for the differential equation: dy/dx = 2x - y. I was flabbergasted to see a differential equations textbook do exactly what I did. Not to sound pretentious, but I really want to know how this is solved. I can't stand having any math be unsolvable. I was researching a bit and is this at all similar to integral of sinx/x?
    I used various forms of Integration by Parts, U-Substitution, Partial Fractions, and a couple others and it was unsuccessful. It looks like a inverse sinx on the bottom, but it's probably a much uglier solution. I think it requires a bit more advanced Calculus with series or something so that's where you guys come in! I would appreciate it if you guys can nudge me in the right direction.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 25, 2009 #2

    Mark44

    Staff: Mentor

    Better get used to it. Most integrals cannot be calculated exactly, so there are lots of techniques for numerical approximation of these integrals.
     
  4. Nov 25, 2009 #3
    So this one is unsolvable in terms of finding an exact solution/function?
     
  5. Nov 25, 2009 #4

    Mark44

    Staff: Mentor

    I don't know, but I would guess that it probably can't be solved analytically.
     
  6. Nov 26, 2009 #5
    Well, after x = sin y, that's just sin(sin(y)) dy which you can expand into a series with respect to powers of y if you really really want to solve it...
     
  7. Nov 26, 2009 #6
    Thanks for the replies guys! Oh, wow, I can't believe I didn't see that before.
     
  8. Nov 26, 2009 #7
    Ugh, this is lame. sin(sinx) has local maxes and local mins at rotating multiples of pi/2. Local maxes are at pi/2 + 2pi(k). Local mins are at 3pi/2 + 2pi(k). It has points of inflection at multiples of pi.
    It looks just like the trig graph of sin(1)sinx. They have the exact same points of inflections and local extrema, yet they differ very slightly in all the other points. If only they were equal to each other.. The integral of sin1sinx is so much easier. I guess sin(sinx) by itself really is impossible to integrate analytically.
    In other words, I got http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/f/d/8fd759fa1ec45ab3f2e3d21d7af57b7f8e76c775.gif [Broken] only when x = multiples of pi/2 or 0. But I think it's impossible to find a function that works for all other points or every point in general.
     
    Last edited by a moderator: May 4, 2017
  9. Nov 26, 2009 #8
    [tex] \sin y = \sum_{n=0}^{\infty} \frac{(-1)^{n} y^{2n+1}}{(2n+1)!} [/tex]

    [tex] \sin(\sin(y)) = \sum_{n,m =0}^{\infty} \frac{(-1)^{n} \left(\frac{(-1)^{m} y^{2m+1}}{(2m+1)!} \right)^{2n+1}}{(2n+1)!} [/tex]

    [tex] \int dy \sin(\sin(y)) = \sum_{n,m =0}^{\infty} \frac{(-1)^{n} \left(\frac{(-1)^{m}}{(2m+1)!} \right)^{2n+1}}{(2n+1)!} \frac{y^{(2m+1)(2n+1)+1}}{(2m+1)(2n+1)}[/tex]

    Enjoy.
     
  10. Nov 26, 2009 #9
    thanks a lot! i love you.
     
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