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Integration/calc work? what did i do wrong

  1. Sep 25, 2005 #1
    Int[1/(x^2+4)^2]
    let x= 2 tan(q)
    i get 1/(4 tan^2(q) + 4)^2
    I get 1/ (16 tan ^4(q) + 32 (1-sec^2(q) + 16)
    and now i think about doing a u-substituition with tan x=u, du =sec^2 (x) can i do that!??!?!?! :surprised :mad:
     
  2. jcsd
  3. Sep 25, 2005 #2
    do you just dont like my topics i want help in?
     
  4. Sep 25, 2005 #3
    [tex] \frac{1}{(4\tan(q)+4)^2} = \frac{1}{16\sec^4q} [/tex]
     
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