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Aggravating Calculus problem; proceed at your own risk :P

  1. Aug 29, 2009 #1
    1. [tex]\int \frac{x}{(x-3)^{2}+1}[/tex]
    which was had by completing the square integrate
    [tex]\int \frac{x}{x^{2}-6x+10}[/tex]

    [tex] u = x[/tex]
    [tex] du = 1[/tex]
    [tex] dv=\frac{1}{(x-3)^{2}+1) }[/tex]
    [tex] v=arctan(x-3)[/tex]

    3. The attempt at a solution
    This problem was presented in a refresher sheet for an engineering course I'm taking this semester. I completed many long and tedious integrations but seem to run into a problem on only this one from the entire sheet. I could easily just forget this problem and use the arctan solution in the integral tables which I inadvertently memorized while trying to solve this problem. If I attempt to integrate by parts from the beginning using the completed square I get the following; which is where I run into a problem. If someone could work out a proper solution I would be eternally grateful even if you just dump it in without using latex.



    [tex]x*arctan(x-3) - \int arctan(x-3)[/tex]

    Integrating arctan when only x or x with a coefficient is easily done by integrating by parts. However I seem to repeat the initial problem when I try to integrate it regardless. The whole difficulty seems to lie with having u=(x-3) rather than (x) or (nx)....
     
    Last edited: Aug 29, 2009
  2. jcsd
  3. Aug 29, 2009 #2

    arildno

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    You may proceed as follows:
    [tex]\int\frac{x}{(x-3)^{2}+1}dx=\int\frac{(x-3)+3}{(x-3)^{2}+1}dx=\int\frac{(x-3)}{(x-3)^{2}+1}dx+\int\frac{3}{(x-3)^{2}+1}dx[/tex]

    Both of these two integrals are easy to compute.
     
  4. Aug 29, 2009 #3
    you sir are made of win sauce!

    I will post my final answer in a moment for anyone else who stumbles upon this problem!
     
  5. Aug 29, 2009 #4
    Wow thanks for that trick, I will make sure to utilize it fully in the future. Silly me!
    Assuming I made no mistakes:

    [tex]=\frac{1}{2}ln((x-3)^{2}+1) + 3arctan(x-3)[/tex]
     
  6. Aug 29, 2009 #5
    And I'm happy to say that Mathematica agrees with me :)
    Thanks, you really rock!
     
  7. Aug 29, 2009 #6

    arildno

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    Okay, I guess I can't complain for lack of compliments...:shy:
     
  8. Sep 14, 2009 #7
    [tex]C_{T}=\int^{0}_{1}\frac{\sigma}{2}r^{2}c_{l}dr[/tex]

    [tex]\sigma=\frac{Nc}{\pi R}[/tex]
    this assumes constant chord (i need to revise this for blade taper)
    N: Number of blades
    c: chord
    R: radius

    [tex]c_{l}=a\alpha[/tex]
    a: rotor slope of leading edge
    alpha: angle of attack

    [tex]C_{p}=\int \lambda dC_{T}+\int^{0}_{1}\frac{\sigma}{2}r^{3}c_{d}dr[/tex]

    [tex]\lambda=\frac{\sigma a}{16}[\sqrt{1+\frac{32}{\sigma a}\theta r}-1][/tex]
     
    Last edited: Sep 14, 2009
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