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Integration reduction formula

  1. Sep 6, 2004 #1
    I'm trying to find an integration reduction formula for the following equation:

    [tex]
    {{I}_n}=\int _{0}^{2}{{\big(4-{x^2}\big)}^n}\delta x
    [/tex]

    Any indication on how to begin would be much appreciated as I've tried many different approaches but all have ended in failure.

    Thanks
     
    Last edited: Sep 6, 2004
  2. jcsd
  3. Sep 6, 2004 #2

    Tide

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    I'd try to repeatedly integrate by parts or possibly use the binomial expansion.
     
  4. Sep 6, 2004 #3

    Zurtex

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    I think I have a solution just give me 5 mins to see if it works.
     
  5. Sep 6, 2004 #4

    Zurtex

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    O.K it's been quite a few months since I've done this, so I can't remember if this reduction formulae is fairly simple.

    If you use the substitution:

    [tex]x = 2 \sin u[/tex]

    It becomes:

    [tex]\frac{1}{2}4^n\int_0^{\frac{\pi}{2}} \left( \cos^{2n-1} u \right) du[/tex]

    I'm sure that can be done with a few trig identities and standard results but it's too late for me to think about it sorry.
     
  6. Sep 6, 2004 #5
    Thanks for the swift responses guys I'll have a go at that tomorrow.
     
  7. Sep 7, 2004 #6

    Zurtex

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    I'm not sure about this but referring back to my previous post could you just let m = 2n - 1 for n > 0 and then that's a fairly standard reduction formulae. I've never done something like that for a reduction formulae but I don't see why it can't be done.
     
  8. Sep 7, 2004 #7
    You may well be able to do that, Zurtex, and I also don't see why it wouldn't work, the only trouble is that it wouldn't prove the relation I was asked to prove.

    I managed to solve it (with help from maths teacher) using a very clever trick indeed. The solution is as follows if anyone is interested:

    [tex]
    {I }_n}\multsp =\int _{0}^{2}{{\big(4-{x^2}\big)}^n}\delta x \\\noalign\vspace{1.08333ex}} \\= {{{{\big[x{{\big(4-{x^2}\big)}^n}\big]}_0}}^2}+2n\int _{0}^{2}{x^2}{{\big(4-{x^2}\big)}^{n-1}}\delta
    x \\\noalign{\vspace{1.08333ex}} \\ \multsp \multsp \multsp \multsp \multsp \multsp =\multsp 2n\int _{0}^{2}\big(4-\big[4-{x^2}\big]\big){{\big(4-{x^2}\big)}^{n-1}}\delta x
    [/tex]

    [tex]
    \noalign{\vspace{1.08333ex}} \\ {{I }_n}\multsp \multsp =\multsp 8n\int _{0}^{2}{{\big(4-{x^2}\big)}^{n-1}}\delta x-2n\multsp {{I }_n} \\ \noalign{\vspace{0.833333ex}}
    [/tex]

    [tex]
    {{I }_n}\multsp \multsp =\multsp 8n\multsp {{I }_{n-1}}-2n\multsp {{I }_n}
    [/tex]

    [tex]
    \noalign{\vspace{0.916667ex}} \\
    {{I }_n}\multsp \multsp =\multsp \frac{8n}{2n+1}{{I }_{n-1}
    [/tex]

    The trick, which I wouldn't have thought of for a very long time, was to write the [tex]x^2[/tex] term as [tex](4-[4-x^2])[/tex]
     
    Last edited: Sep 7, 2004
  9. Sep 7, 2004 #8

    Zurtex

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    I thought I'd seen that before, that's really silly of me not to spot. Well done for working it out.
     
  10. Apr 1, 2009 #9
    What happens in the very first step of the solution?
     
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