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Reduction formula

  1. Jan 30, 2014 #1
    If [itex]I_{n}=\int_0^1 (1-x^{3})^{n} dx[/itex], use integration by parts to prove the reduction formula [itex]I_{n}=\frac{3n}{3n+1}I_{n-1}[/itex]


    My attempt: let [itex]u=(1-x^{3})^{n}[/itex], and [itex]dv=dx[/itex]. Then [itex]I_{n}=[(1-x^{3})^{n}x]_0^1 - \int_0^1 -3x^{2}n(1-x^{3})^{n-1}x dx = 3n \int_0^1 x^{2}(1-x^{3})^{n-1} dx[/itex]. But I don't know where to go from here. Any help would be appreciated.
     
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  3. Jan 30, 2014 #2

    Dick

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    You have an x^3 in your last integral. Not an x^2. Here's a hint. Try writing x^3=(1-x^3)-1.
     
  4. Jan 30, 2014 #3
    After writing x^3=(1-x^3)-1, should I integrate by parts again?
     
  5. Jan 30, 2014 #4

    Dick

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    I think you can think of something cleverer than that. Split it into two integrals and take a close look at them.
     
  6. Jan 30, 2014 #5

    Dick

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    Ooops. I've got a typo. That should obviously be x^3=(x^3-1)+1. Sorry!
     
  7. Jan 30, 2014 #6
    I've got it now. Thanks :D
     
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