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Integration Help (Quick if Possible!)

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int sec^2 x tan x dx[/tex]

    2. Relevant equations



    3. The attempt at a solution
    hey guys, sorry about this but i realised i can't do this question (in for tommorow)

    I said, let u = tan x
    so du/dx = sec^2 x
    so dx/du =1 / sec^x

    so putting that in I get [tex]\int tan x dx or \int u du[/tex]

    which of course is [tex]\frac{u^2}{2}[/tex] = [tex]\frac{tan^2 x}{2}[/tex]

    however according to wolfram alpha it's sec^2 x / 2 :O

    http://www.wolframalpha.com/input/?i=int+sec^2+x+tan+x

    thanks again in advance, would save me :O
     
  2. jcsd
  3. Oct 13, 2009 #2

    Mark44

    Staff: Mentor

    Your answer is correct. Another substitution you can use is u = secx, du = secx tanx dx. With that substitution you have
    [tex]\int u du~=~(1/2)u^2 + C~=~(1/2)sec^2x + C[/tex]
    The two answers seem to be different, and indeed they are. But they differ only by a constant, since (1/2)sec^2(x) = (1/2)tan^(x) + 1/2.

    The upshot is that you are correct and wolfram alpha is also correct.
     
  4. Oct 13, 2009 #3
    i see now, much obliged sir!
     
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