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Trig. integration tan^5(x)dx

  1. Oct 3, 2009 #1
    I'm not sure if my answer is just wrong or basically the same as the one in the back of my book.
    1. The problem statement, all variables and given/known data
    [tex]\int tan^5(x)dx[/tex]

    3. The attempt at a solution
    My answer:

    [tex]\int tan^5(x)dx = \frac{tan^4(x)}{4} - \int(sec^2(x)tan(x) - tan(x)) dx[/tex]

    [tex]\int tan^5(x)dx = \frac{tan^4(x)}{4} - \frac{tan^2(x)}{2} + ln|sec(x)| + C[/tex]

    Book answer:

    [tex]\int tan^5(x)dx = \frac{sec^4(x)}{4} - tan^2(x) + ln|sec(x)| + C[/tex]

    If I understand correctly,

    [tex]\frac{tan^4(x)}{4} + C = \frac{sec^4(x) - 1}{4} + C = \frac{sec^4(x)}{4} + C[/tex]

    If that's wrong let me know please.

    However, I don't get why I keep getting

    [tex]\frac{tan^2(x)}{2}[/tex]

    instead of

    [tex]tan^2(x)[/tex]
     
  2. jcsd
  3. Oct 3, 2009 #2

    Dick

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    Ooops. tan(x)^4=(sec(x)^2-1)^2. Multiply the square out.
     
  4. Oct 3, 2009 #3
    Ah! So that means my initial answer was right (assuming I don't keep screwing up my trig), since:

    = tan^4(x)/4 - tan^2(x)/2 + ... + C

    = [sec^4(x) - 2sec^2(x) + 1]/4 - tan^2(x)/2 + ... + C

    = sec^4(x)/4 - sec^2(x)/2 - tan^2(x)/2 + ... + C

    = sec^4(x)/4 - [tan^2(x) + 1]/2 - tan^2(x)/2 + ... + C

    = sec^4(x)/4 - tan^2(x) + ... + C

    Thanks
     
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