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porkstein Offline Male 

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Last Activity 23-08-2013

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  1. Psh, look at all these people trying to find integrals. Look at what I did with your sig.
  2. Yep!!!...
    ...okay maybe not. But not doing either action can be non-contradictory :P . One can simply feel too lazy to be bothered enough to pull the leaver.
  3. I'd not pull the lever to stop it killing one person because I'm lazy. I'd not pull it to stop it killing five people because of overpopulation...
    :awesome: .
  4. I used parts on (sec x . tan x) . tan x and got the same answer.
  5. I got the same!
  6. Is the answer to the integral in your sig

    \frac{1}{2}(secxtanx - ln(secx + tanx)) + C ?

    (Or some variation thereupon?).

    If it is, I basically just used the fact it becomes cyclical by separating it into two integrals of sec^3x and secx and then by using the integral of sec^3x I used used IBP twice and took the second integral of sec^3x to the other side, divided by two and minused the integral of secx from it.

    Was a fun little question.

    (If it isn't right, :facepalm: ).
  7. If I could neg you for having a stupid signature I would, but alas I have used everything up.
  8. The maths equation in your sig - is it actually possible?

About Me

  • About porkstein

    Star Sign
    Hydrogen wave function:

    \displaystyle \psi(r, \theta, \phi) =  \sqrt{\left( \frac{2}{n} \right) ^3 \frac{(n-l-1)!}{2n(n+l)!}}L^{2l+1}_{n-l-1}\left( \frac{2r}{n} \right)  e^{-r/n} \left( \frac{2r}{n} \right) ^l \cdot \sqrt {\frac{(2l+1)(l-m)!}{4 \pi (l+m)!}} P_l^m(\cos\theta) e^{im\phi}


    \displaystyle L^a_b(x) = \frac{x^{-a} e^x}{b!} \frac{d^b}{dx^b} \left(e^{-x} x^{a+b} \right)

    \displaystyle P^a_b(x) = \frac{(-1)^a}{2^b b!} \left(1-x^2\right)^{a/2} \frac{d^{a+b}}{dx^{a+b}}\left(x^  2 - 1\right)^b

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  • Last Activity 23-08-2013
  • Join Date 13-01-2009

Location UK

Join Date 13-01-2009

Total Posts 980

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