Results 1 to 4 of 4

Math Help - Integral proof

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    133

    Integral proof

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

     \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 u=(x^2+a^2)^n \\, dv=dx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    This looks like a reduction formula.

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

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

    Using integration by parts:

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

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

    Hence 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

    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.....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    41
    Quote Originally Posted by Showcase_22 View Post
    This looks like a reduction formula.

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

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

     \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)

    u = x\; \text{and}\; dv = x(x^2+a^2)^{n-1}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Talking

    <br /> <br />
\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<br />
    <br /> <br />
\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<br />

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

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

    \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

    \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

    \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

    \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

    \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

    \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!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Integral proof ..
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 15th 2010, 03:25 PM
  2. Integral Proof
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 14th 2010, 03:29 AM
  3. integral/sin proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 17th 2010, 03:53 PM
  4. Help with Integral proof
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 21st 2009, 06:09 PM
  5. Replies: 6
    Last Post: August 4th 2007, 09:48 AM

Search Tags


/mathhelpforum @mathhelpforum