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N-dim integral

  1. Sep 17, 2011 #1
    Hello

    I have problem with integrate

    [tex]
    \int_{0}^{\frac{\pi}{2}}d\phi_1...\int_{0}^{\frac{\pi}{2}}d\phi_{n-2}sin^{2(n-1)}\phi_1...sin^{2(n-k)}\phi_k ...sin^4\phi_{n-2}
    [/tex]

    Please help me.
     
  2. jcsd
  3. Sep 17, 2011 #2
    Ok, you got to start small with these things then build it back up. So how about starting with n=2:

    [tex]\int_0^a\int_0^a \sin^2(\phi_1)\sin^4(\phi_2) d\phi_1 d\phi_2[/tex]

    Alright, get that one straight, then add another one, then another one. Do maybe 3,4 or five that way and you'll (hopefully) see a trend that you can then use deduction to deduce the value for the general expression.
     
  4. Sep 17, 2011 #3
    Using the Mathematica I found that
    [tex]\int_{0}^{\frac{\pi}{2}}Sin^2xdx=\frac{\pi}{4}[/tex][tex]\int_{0}^{\frac{\pi}{2}}Sin^4xdx=\frac{3\pi}{16}[/tex]
    [tex]\int_{0}^{\frac{\pi}{2}}Sin^6xdx=\frac{5\pi}{32}[/tex][tex]\int_{0}^{\frac{\pi}{2}}Sin^8xdx=\frac{35\pi}{256}[/tex]
    [tex]\int_{0}^{\frac{\pi}{2}}Sin^{10}xdx=\frac{63\pi}{512}[/tex]

    so I can write

    [tex]\int_{0}^{\frac{\pi}{2}}Sin^{2n}xdx=\frac{a_{2n} \pi }{2^{2n}}[/tex]

    but I cant find any logical expression which describes an
     
  5. Sep 17, 2011 #4
    Keep in mind the variables are separated so won't the answer be some kind of product like:

    [tex]\displaystyle\prod_{n=1}^N (I_n)[/tex]

    I think so anyway. So the [itex]a_n[/itex] term may be a problem. Ok, what happens when you just solve for the antiderivative when n=2, n=3, n=4, n=5. Can you see some kind of trend there?.
     
  6. Sep 17, 2011 #5
    I think I may have led you astray on this and now suggest we focus on the recursive formula:

    [tex]\int \sin^n(x)dx=-\frac{1}{n} \cos(x)\sin^{n-1}(x)+\frac{n-1}{n}\int \sin^{n-2}(x)dx[/tex]

    Haven't worked it out yet but it looks encouraging.
     
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