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Homework Help: Problem with u-Substitution

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex]tan x sec[tex]^{}4[/tex] x dx

    2. Relevant equations

    3. The attempt at a solution
    I did this two different ways and go two different answers.

    The correct way is
    [tex]\int[/tex]sec^3(x)*sec(x)*tan(x) dx
    letting u= sec(x) and du= sec(x)*tan(x)
    = [tex]\int[/tex] u^3 du = sec^4(x)/4 + C
    Which is good,

    but I can't figure out what's wrong with this way:
    [tex]\int[/tex]tan(x)*sec^4(x) dx
    = [tex]\int[/tex] tan(x)*(1+tan^2(x))(sec^2(x)) dx
    from the identity sec^2(x) = 1 + tan^2(x)
    then letting u= tan(x) and du= sec^2(x) dx
    gives [tex]\int[/tex] u*(1 + u^2) du = [tex]\int[/tex] u + u^3 du = u^2/2 + u^4/4 + C
    = tan^2(x)/2 + tan^4(x)/4 + C

    Which is a different answer from the correct one. What's wrong with the second way? Does it have to do with the u-sub... Thanks,
    Last edited: Jun 10, 2009
  2. jcsd
  3. Jun 10, 2009 #2


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    Homework Helper

    Both answers are correct, but the constant C in your second answer isn't the same as the constant C in your first answer.

    \tan^2(x)/2 + \tan^4(x)/4+C'=\frac{1}{4}(\sec^4(x)-1)+C'=\sec^4(x)/4-1/4+C'=\sec^4(x)+A

    I leave the intermediate steps to you.
  4. Jun 10, 2009 #3


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    Homework Helper


    [tex]sec^4x= 1 + 2tan^2x+tan^4x[/tex]

    Now divide by 1/4 and you'll see that they are the exact same thing.

    AND 1/4+C = another constant!
  5. Jun 10, 2009 #4
    Thanks a lot!
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