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Integral of tan(x)^3

  1. Mar 3, 2009 #1
    I am having an issue with this problem.

    [tex]\int tan(x)^3: [/tex]

    You separate a tan(x)^2, and use the identity 1+tan(x)^2 = sec(x)^2

    You then end up with [tex]\int tan(x) sec(x)^2 dx[/tex] + [tex]\int tan(x) dx[/tex]

    [tex]\int tan(x) dx = ln absval(sec(x)) [/tex]

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

    Here, the book says to solve the problem this way:

    set u = tan(x) du = sec(x)^2dx

    so [tex]\int u du = 1/2 u ^2 = 1/2 tan(x)^2 [/tex]

    Why can't you solve this problem this way instead?

    [tex]\int (tan(x) sec(x)) sec(x) dx[/tex]

    set u = sec(x), du = (tan(x) sec(x)) dx

    so [tex]\int u du = 1/2 u ^2 = 1/2 sec(x)^2 [/tex]

    I think I must be missing something...any help you could give me would be greatly appreciated.
     
  2. jcsd
  3. Mar 3, 2009 #2

    Dick

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    You can solve it that way. What you are missing is the +C. An indefinite integral always has an indefinite integration constant. (1/2)*tan(x)^2+C and (1/2)*sec(x)^2+D are the same thing. tan^2 and sec^2 differ by a constant. Only the C and D are different constants, right?
     
  4. Mar 3, 2009 #3
    you arent doing anything wrong. both answers are correct. they just differ by a constant. Remember

    sec(x)^2 - tan(x)^2 = 1
     
  5. Mar 3, 2009 #4
    Thanks for replying so quick.

    If it is a definite integral don't the constants cancel out?
     
  6. Mar 3, 2009 #5

    Dick

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    Yes, that's the point. (1/2)tan(x)^2+C and (1/2)sec(x)^2+C both differentiate to the same thing. They are both antiderivatives of tan(x)*sec(x)^2. If you try to do a definite integral, you don't care what which form you use. You get the same thing.
     
  7. Mar 3, 2009 #6
    I see what you are saying, this makes sense to me now...thanks a lot for getting back to me so quick, I really appreciate it.
     
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