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Integrating 1/(Z^2 + A^2) - something wrong?

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{1}{2t^2+4} \mathrm{d}t[\tex]


    2. Relevant equations
    [tex]\int \frac{1}{Z^2+A^2} \mathrm{d}Z = \frac{1}{A} \arctan{(\frac{Z}{A})} + c[/tex]


    3. The attempt at a solution
    Looks quite easy, but this is what's annoying me: the two methods below should be identical, but something's gone wrong and I can't work out what. For some reason, the second method gives an extra factor of 1/sqrt(2) at the front. Why aren't they both the same?

    [tex]\int \frac{1}{2t^2+4} \mathrm{d}t = \int \frac{1}{(\sqrt{2t})^2+2^2} \mathrm{d}t = \frac{1}{2} \arctan {(\frac{\sqrt{2}t}{2})} + c[/tex]

    [tex]\int \frac{1}{2t^2+4} \mathrm{d}t = \frac{1}{2} \int \frac{1}{t^2+2} = \frac{1}{2 \sqrt{2}} \arctan{(\frac{\sqrt{2} t}{2})}+c[/tex]

    With very many thanks,

    Froskoy.
     
  2. jcsd
  3. Dec 16, 2011 #2
    Are you sure that c represents the same value in each?
     
  4. Dec 16, 2011 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    You forgot to use [itex] \sqrt{2} dt[/itex] in the first integral.

    RGV
     
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