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Thread: Integral proof

  1. #1
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    Integral proof

    In my sleep deprived state, I cant figure out the correct substitute to make in order to get this proof to work;

    $\displaystyle \int (x^2+a^2)^n \,dx = \frac{x(x^2+a^2)^n}{2n+1}+\frac{2na^2}{2n+1}\int (x^2+a^2)^{n-1}\,dx$

    I was using $\displaystyle u=(x^2+a^2)^n \\, dv=dx$
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  2. #2
    Super Member Showcase_22's Avatar
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    This looks like a reduction formula.

    Let $\displaystyle I_n=\int (x^2+a^2)^n~dx$

    $\displaystyle I_n=\int(x^2+a^2)^{n-1} (x^2+a^2)~dx$

    Using integration by parts:

    $\displaystyle u=(x^2+a^2)^{n-1} \Rightarrow \ \frac{du}{dx}=2x(n-1)(x^2+a^2)^{n-2}$

    $\displaystyle \frac{dv}{dx}=x^2+a^2 \Rightarrow \ v=\frac{x^3}{3}+a^2 x$

    Hence $\displaystyle I_n=\left( \frac{x^3}{3}+a^2x \right) (x^2+a^2)^{n-1}-\int \left( \frac{x^3}{3}+a^2 x \right) (2x)(n-1)(x^2+a^2)^{n-2}~dx$

    $\displaystyle I_n=x \left( \frac{x^2}{3}+a^2 \right)(x^2+a^2)^{n-1}-2(n-1) \int x^2 \left( \frac{x^2}{3}+a^2 \right) (x^2+a^2)^{n-2}~dx$

    hmmm, i'll keep trying with this.....
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    This looks like a reduction formula.

    Let $\displaystyle I_n=\int (x^2+a^2)^n~dx$

    $\displaystyle I_n=\int(x^2+a^2)^{n-1} (x^2+a^2)~dx$
    The split is good.

    $\displaystyle \int(x^2+a^2)^{n-1} (x^2+a^2)~dx = \int x^2 (x^2+a^2)^{n-1} ~dx + a^2 \int (x^2+a^2)^{n-1} ~dx$

    and on the first integral try (by parts)

    $\displaystyle u = x\; \text{and}\; dv = x(x^2+a^2)^{n-1}$
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  4. #4
    Super Member Showcase_22's Avatar
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    $\displaystyle

    \int (x^2+a^2)^n \,dx = \frac{x(x^2+a^2)^n}{2n+1}+\frac{2na^2}{2n+1}\int (x^2+a^2)^{n-1}\,dx
    $
    $\displaystyle

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

    Let $\displaystyle u=x \Rightarrow \ \frac{du}{dx}=1$

    $\displaystyle \frac{dv}{dx}=x(x^2+a^2)^{n-1} \Rightarrow \ v=\frac{1}{2n}(x^2+a^2)^n$

    $\displaystyle \int(x^2+a^2)^{n-1} (x^2+a^2)~dx =\frac{x}{2n}(x^2+a^2)^n-\frac{1}{2n} \int (x^2+a^2)^n~dx+a^2 \int (x^2+a^2)^{n-1}~dx$

    $\displaystyle \Rightarrow \ I_n=\frac{x}{2n}(x^2+a^2)^n-\frac{1}{2n}I_n+a^2 \int (x^2+a^2)^{n-1}~dx$

    $\displaystyle \Rightarrow I_n \left( 1+\frac{1}{2n} \right)=\frac{x}{2n}(x^2+a^2)^n+a^2 \int (x^2+a^2)^{n-1}~dx$

    $\displaystyle \Rightarrow I_n \left( \frac{2n+1}{2n} \right)=\frac{x}{2n}(x^2+a^2)^n+a^2\int (x^2+a^2)^{n-1}~dx$

    $\displaystyle \Rightarrow I_n=\frac{x}{2n+1}(x^2+a^2)^n+\frac{2na^2}{2n+1}\i nt (x^2+a^2)^{n-1}~dx$

    $\displaystyle \Rightarrow I_n=\frac{x(x^2+a^2)^n}{2n+1}+\frac{2na^2}{2n+1}\i nt (x^2+a^2)^{n-1}~dx$

    YAY!!!
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