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Homework Help: Is my book wrong?

  1. Apr 14, 2005 #1

    mad

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    I have this integration problem that I did but it doesnt give me the right answer. But there are like 3 other similar exercices I did the same way and I got all the right answers.. maybe it is my book (I dont think so ... :p)

    It's an integration problem:

    [tex]\int tg^3(4x) sec^4(4x)dx[/tex]

    heres what I did:

    [tex]\int (sec^2(4x)-1)(tg(4x))(sec^4(4x))dx[/tex]
    = [tex]\int (sec^6(4x) - sec^4(4x)) (tg4x) dx [/tex]

    u= sec 4x
    du = 4(sec4x)(tg4x)dx --> dx = du/(4(sec4x)(tg4x))

    =[tex]\int \frac{(u^6 - u^4) du}{4u} [/tex]

    (replaced the sec(4x) at denom. with u since u=sec4x)

    =[tex]\frac{1}{4}\int u^5 - [/tex] [tex]\frac{1}{4}\int u^3 [/tex]
    = [tex]\frac{1}{24}sec^6(4x) -[/tex] [tex]\frac{1}{16} sec^4(4x) [/tex] +C


    Add anything you want! Thanks everyone

    BTW the answer in my book is
    (1/16) tg^4 (4x) + (1/24) tg^6(4x)
    I tried it in my calc with an x and it doesnt give the same answer.
     
    Last edited: Apr 14, 2005
  2. jcsd
  3. Apr 14, 2005 #2

    Imo

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    Actually, those aer the same answers (up to that constant, if you put it directly into your calculator, they differ already by a constant that doesn't matter in this type of integration...add 1/48 to your answer to get the other). They just went another direction in the integration.
     
  4. Apr 14, 2005 #3
    What the hell is tg(x)?
     
  5. Apr 14, 2005 #4

    dextercioby

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    Here's another way.Denote the integral by I and write everything in terms of sine a cosine...

    [tex] I=\int \left(\frac{\sin^{3}4x}{\cos^{7}4x}\right) \ dx [/tex]

    [tex] 4x=u \Rightarrow 4dx=du [/tex]

    [tex] I=\frac{1}{4}\int \left(\frac{\sin^{4}u}{\cos^{7}u}\right) \ du
    =-\frac{1}{4}\int \left(\frac{1-\cos^{2}u}{\cos^{7}u}\right) \ d(\cos u) =-\frac{1}{4}\left[\frac{(\cos u)^{-6}}{-6}-\frac{(\cos u)^{-4}}{-4}\right] +\mathcal{C}[/tex]

    Therefore,reversing the substitution made

    [tex]I=\frac{1}{24}\frac{1}{\cos^{6}4x}-\frac{1}{16}\frac{1}{\cos^{4}4x}+\mathcal{C} [/tex]

    Daniel.
     
  6. Apr 14, 2005 #5

    dextercioby

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    Tangent of (x),what else??? :rolleyes:

    Daniel.
     
  7. Apr 14, 2005 #6

    mad

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    Thanks for your help, Daniel.
    I see my solution was okay. We had to do the problem by exponents of sec and tg. (sorry I dont know what the method is called in english)

    However, what does d(cos u) means in your solution? and I know you replaced a sin^4 (u) by (1-cos^2 (u)), but where is the other sin^2 (x)
    Surely it is the d(cos u) you used, but I'm not familiar with this notation
     
    Last edited: Apr 14, 2005
  8. Apr 14, 2005 #7
    What happened to tan(x)?
     
  9. Apr 14, 2005 #8

    cepheid

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    Skin cancer you know...the poor devil.
     
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