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Integration for BC Calculus

  1. Feb 12, 2006 #1
    I broke up the inegral of tan^(7)(theta)*sec^(5)(theta) into tan^(5)(theta)(sec^(2)(theta))(sec^(5)(theta). WHAT DO I DO NEXT
     
  2. jcsd
  3. Feb 12, 2006 #2

    arildno

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    You have broken it up incorrectly.
     
  4. Feb 13, 2006 #3

    VietDao29

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    As arildno have pointed out, you've broken it incorrectly.
    [tex]\tan ^ 5 \theta \sec ^ 2 \theta \sec ^ 5 \theta = \tan ^ 5 \theta \sec ^ 7 \theta \neq \tan ^ 7 \theta \sec ^ 5 \theta[/tex]
    In this integral, by converting tangent function to sine, and cosine function, we have:
    [tex]\int \tan ^ 7 \theta \sec ^ 5 \theta d \theta = \int \frac{\sin ^ 7 \theta}{\cos ^ {12} \theta} d \theta[/tex]
    Now the power of the sine function is odd, it's common to use the substitution: [tex]u = \sin \theta[/tex], then use the well-known Pythagorean identity: cos2x + sin2x = 1 (or you can rearrange it a bit to give: sin2x = 1 - cos2x), to solve the problem.
    If the power of the cosine function is odd, then it's common to use the substitution: [tex]u = \cos \theta[/tex].
    Can you go from here? :)
     
    Last edited: Feb 13, 2006
  5. Feb 13, 2006 #4

    Hurkyl

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    The textbooks I've seen usually present an algorithm for doing tan * sec integrals directly.

    (That's as much pointed at nyyfan0729 as it is at VietDao29)
     
  6. Feb 13, 2006 #5

    arildno

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    Asa follow-up on Hurkyl's suggestion, remember that:
    [tex]\frac{d}{dx}\tan(x)=\sec^{2}(x)[/tex]
     
  7. Feb 13, 2006 #6

    Hmmm, I guess I'm missing something because to me it seems like it's more important to remember that [tex]\frac{d}{dx}\sec(x)=\sec(x)tan(x)[/tex]

    Also if you do this with sine & cosine, as VietDao suggested, then I'm pretty sure that you need to make [tex]u = \cos \theta[/tex] the substitution instead of [tex]u = \sin \theta[/tex]
     
    Last edited: Feb 13, 2006
  8. Feb 13, 2006 #7

    arildno

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    Well, if you set [tex]\frac{dv}{dx}=\tan^{7}(x)\sec^{2}(x)[/tex]
    then the original integral is easily integrated as follows:
    [tex]\int\tan^{7}(x)\sec^{5}(x)dx=\frac{1}{8}\tan^{8}(x)\sec^{3}(x)-\frac{3}{8}\int\tan^{9}(x)\sec^{3}(x)dx=\frac{1}{8}\tan^{8}(x)\sec^{3}(x)-\frac{3}{80}\tan^{10}(x)\sec(x)+\frac{1}{120}\tan^{12}(x)+C[/tex]
    or something like that.

    Hmm..did a make a mistake somewhere?

    Aargh, seems that I did..:grumpy:
     
    Last edited: Feb 13, 2006
  9. Feb 14, 2006 #8

    VietDao29

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    Yes, thanks, :blushing:
    Whoops, sorry. My bad... :blushing:
    What the hell was I thinking about when writing this??? :grumpy: :grumpy: :grumpy:
     
  10. Feb 14, 2006 #9

    GCT

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    it seems that CrankFan has a nice suggestion, try setting [tex]u=sec \theta [/tex] and simplifying it from there.
     
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