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Math Help - Use Integration by Parts to prove the Reduction Formula

  1. #1
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    Use Integration by Parts to prove the Reduction Formula

    int[(x^2 + a^2)^n] dx =
    [x(x^2 + a^2)/(2n + 1)] + [(2na^2)/(2n + 1)]*int[(x^2 + a^2)^(n-1)] dx

    Where n =/= -(1/2)

    I've done a few other examples of integration by parts just fine, but I can't seem to find selections for u and dv which allow me to solve this.

    If anyone can grant me the FIRST step, I'm sure I can take it from there.

    Cheers!
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  2. #2
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    Quote Originally Posted by Tulki View Post
    int[(x^2 + a^2)^n] dx =
    [x(x^2 + a^2)^n/(2n + 1)] + [(2na^2)/(2n + 1)]*int[(x^2 + a^2)^(n-1)] dx

    Where n =/= -(1/2)

    I've done a few other examples of integration by parts just fine, but I can't seem to find selections for u and dv which allow me to solve this.

    If anyone can grant me the FIRST step, I'm sure I can take it from there.

    Cheers!
    Hi Tulki,

    You can let u=(x^2+a^2)^n,\ dv=dx,\ v=x

    Or let dv=1, v=x.

    Integration by parts gives

    \int{\left(x^2+a^2\right)^n}dx=x\left(x^2+a^2\righ  t)^n-\int{nx\left(x^2+a^2\right)^{n-1}(2x)}dx

    =x\left(x^2+a^2\right)^n-2n\int{\left(x^2-a^2+a^2\right)\left(x^2+a^2\right)^{n-1}}dx

    (Adding a^2-a^2 to x^2 helps solve the integral easily).

    =x\left(x^2+a^2\right)^n-2n\left[\int{\left(x^2+a^2\right)\left(x^2+a^2\right)^{n-1}}dx-\int{a^2\left(x^2+a^2\right)^{n-1}}dx\right]

    =x\left(x^2+a^2\right)^n-2n\int{\left(x^2+a^2\right)^n}dx+2na^2\int{\left(x  ^2+a^2\right)^{n-1}}dx

    \Rightarrow\ I=x\left(x^2+a^2\right)^n-2nI+2na^2I_{n-1}

    \Rightarrow\ I(1+2n)=x\left(x^2+a^2\right)^n+2a^2nI_{n-1}

    \Rightarrow\ I=\frac{x\left(x^2+a^2\right)^n}{1+2n}+\frac{2a^2n  }{1+2n}\int{\left(x^2+a^2\right)^{n-1}}dx
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  3. #3
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    Um... when you quoted the question, you added in a power of n that isn't in the actual question I'm looking at. May have been a mistake, but I'm not sure.
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  4. #4
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    I think it's a misprint.
    I wasn't able to get the answer quoted, Tulki.
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  5. #5
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    Yes, it's definately a misprint, Tulki.
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  6. #6
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    Yeah I've been trying to figure out a way through it as it is printed, but it seems it is a misprint. Thank you!
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