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Math Help - Reduction Method Question

  1. #1
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    Reduction Method Question

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  2. #2
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    First off, there's a typo (an n is missing)

    \int(x^2+a^2)^ndx if u = (x^2+a^2)^n,\;\; dv= dx so du = 2 n x (x^2+a^2)^{n-1},\;\;v = x so

    I_n = \int(x^2+a^2)^ndx
     = x(x^2+a^2)^n - 2n \int x^2 (x^2+a^2)^{n-1}dx
     = x(x^2+a^2)^n - 2n \int (x^2+a^2-a^2) (x^2+a^2)^{n-1}dx
     = x(x^2+a^2)^n - 2n \int (x^2+a^2) (x^2+a^2)^{n-1}dx + 2na^2 \int (x^2+a^2)^{n-1}dx
     I_n = x(x^2+a^2)^n - 2n \underbrace{\int (x^2+a^2)^{n}dx}_{I_n} + 2na^2 \underbrace{\int (x^2+a^2)^{n-1}dx}_{I_{n-1}}
    so

     (1+2n) I_n = x(x^2+a^2)^n +2na^2 I_{n-1}

    from we obtain your result.
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  3. #3
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    Thank you for your help but how did you know their was a typo. I though I had to manipulate the x^2 + a^2 to get it into a high power like saying x^2 + a^2 = (x^2 + a^2) ^n * (x^2 + a^2)^1-n
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  4. #4
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    Quote Originally Posted by shonenhikada View Post
    Thank you for your help but how did you know their was a typo. I though I had to manipulate the x^2 + a^2 to get it into a high power like saying x^2 + a^2 = (x^2 + a^2) ^n * (x^2 + a^2)^1-n
    You circled a term and called it I_{n-1} (I change it for this conversation). This suggests that there is an I_{n}. Second, put a=0,\;\;n=2 - your formula doesn't work!
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