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Should be easy, but I got stuck!

  1. Aug 25, 2008 #1
    Completely stuck with this:

    [tex]\int_0^\infty sin^{2008} x dx[/tex]

    Any help on how to tackle this please?
  2. jcsd
  3. Aug 25, 2008 #2


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    What have you tried? (And are you sure you wrote it correctly?)
  4. Aug 25, 2008 #3
    Thanks for you prompt reply,

    Yes it's written correctly. I've tried a formula that I found in a book, that says it is:

    1*3*5*...*(n-1)/2*4*6*...*(n) * pi

    But it just can't be as simple as that... I'm sure this is a tricky question.
  5. Aug 25, 2008 #4


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    Hmm, I have to say that doesn't seem particular plausible. Think of the area under the graph of [tex]\sin^{2008} x[/tex] from 0 to [tex]\infty[/tex]. Did you mistype the integral limits?
  6. Aug 25, 2008 #5
    Ok, I have re-checked it, I had it wrong, it is from 0 to pi. Sorry guys.

    [tex]\int_0^\pi sin^{2008} x dx[/tex]

    Still buffed though!
  7. Aug 25, 2008 #6
    Hmm well you could apply the usual methods for starting these integrals. Consider breaking down the integral into two (one with limits of integration of 0 to pi/2 and another of pi/2 to pi). Then consider the symmetry of sin(x) to realize that these two integrals are equal. Then try a substitution involving the identity cos(x) = sin(pi/2 - x).

    But then again in this case it might complicate the problem. Is there an easy way to factor sin2008(x) + cos2008(x)
  8. Aug 26, 2008 #7
    I would suggest you try integration by parts to derive a recursion formula for

    I_n=\int_0^\pi{dx \sin^nx}

    Then you can solve this recursion formula and plug in n=2008.
  9. Aug 26, 2008 #8
    I have seen that recursion formula before and I can tell you without remembering the specifics that 2008 is a high enough number that this approach will probably be not very pleasant.
  10. Aug 26, 2008 #9

    I can tell you that the result is not very pleasant:smile:
  11. Aug 26, 2008 #10
    Haha, well I didn't mean to be rude, since my approach seemed to get no where. I forgot that [tex]a^n + b^n[/tex] cannot really be factored readily.
  12. Aug 26, 2008 #11


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    [tex]=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ \int_0^{\pi}\,(2n-1)\,sin^{2n-2}x\,cos^2xdx[/tex]

    [tex]=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ [sin^{2n-1}x\,cosx]_0^{\pi}\ -\ \int_0^{\pi}\,sin^{2n}x\,dx[/tex]

    So [tex]\int_0^{\pi}\,sin^{2n}x\,dx\ =\ \frac{2n-1}{2n}\,\int_0^{\pi}\,sin^{2n-2}x\,dx[/tex] :smile:
  13. Aug 27, 2008 #12
    Yes, and the same works if you take n instead of 2n.:smile:
  14. Aug 27, 2008 #13


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    probability …

    Is [tex]\frac{1}{\pi}\,\int_0^\pi sin^{2008} x dx[/tex]

    the probability of tossing a coin 2008 times :rolleyes: and getting exactly 1004 heads?

    If so, why … ? :confused:

    ([tex]\int_0^\pi sin^m x dx[/tex] is rational for odd m, and a rational multiple of π for even m)
  15. Aug 27, 2008 #14
    How can it be? Tossing a coin is a discrete even -> discrete probability distribution function but integrals are analagous to continous ones. Or am I misunderstanding your question tiny-tim?
  16. Aug 27, 2008 #15
    Well, maybe, Tiny-Tim just observed that the two quantities - the integral on the one hand, and the describes probability - are the same.

    The probability of having k heads in 2k tosses is


    On the other hand, from the recursion formula described above one finds (using [itex]I_0=\pi[/itex])


    Now one can use the identites [itex](2k)!!=2^k k![/itex] and [itex](2k-1)!!=\frac{(2k)!}{2^k k!}[/itex] to find

    \frac{1}{\pi}I_{2k} = \frac{(2k)!}{2^kk!2^kk!}=p_{2k}

    If the question were not about 2008 but some odd number, something similar would probably be true.
  17. Aug 27, 2008 #16


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    Yes … but why?

    :cry: what does it all mean? :cry:
  18. Aug 27, 2008 #17
    I don't know ....
  19. Aug 27, 2008 #18
    Wow, thanks for all the answers!

    I have done my research and from various books I have found the same answer as this. This is all fine, but I have a loooooooong way to reach the number 2008... So I am still confused.
  20. Aug 27, 2008 #19
    See posts #7 and #15.
  21. Aug 27, 2008 #20


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    Hi siaosi! :smile:

    The best thing to do, if you're confused by complicated equations, is to give things short names so that equatins look really simple.

    In this case, define P2n = ∫0π sin2nx dx.

    Then that equation was P2n =
    P2(n-1) (2n-1)/2n,​

    which you can then see is
    P2(n-2) (2n-1)(2n-3)/2n(2n-2),​

    and so on … until you reach P0, which you can easily work out! :smile:

    Now can you see that there's a pattern … so you don't have to worry about making thousands of calculations! :wink:
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