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Another quick integration q

  1. Aug 1, 2008 #1
    the question is [tex]\int 2x sec^2 (x^2) dx[/tex]

    do i sub u= sec (x^2) ?

    I so far have got to trying to sub u= sec(x^2) and getting du= 2 (sec x^2 tan x^2)... i have a strong feeling i fudged the "du" part. hmm.
  2. jcsd
  3. Aug 1, 2008 #2

    [tex]du=2xdx[/tex] which you have! What is the anti-derivative of [tex]\sec^2 x[/tex] ???
  4. Aug 1, 2008 #3
    oh right, so i've got du=2xdx,

    intergral sec^2 u du

    = tan u + c

    sub back u=x^2

    =tan x^2 +c

    how does that look?
  5. Aug 1, 2008 #4


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    It looks just fine.
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