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Trigonometric Integral

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data


    [tex]\int[/tex] 41(sin[tex]^{6}[/tex](x))(cos[tex]^{3}[/tex](x))


    2. Relevant equations



    3. The attempt at a solution

    I think you are supposed to use the half angle identities and then maybe integration by parts but Im lost on it.
     
  2. jcsd
  3. Oct 19, 2009 #2

    lanedance

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    how about starting by trying the substitution u = sinx
     
  4. Oct 19, 2009 #3
    I have tried a few different things, more than I wanted to list but it just keeps getting more and more convoluted. I need a walk through we just started this in class and my teacher doesnt answer questions so Im just a bit lost over all.
     
  5. Oct 19, 2009 #4

    Dick

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    You don't have to do anything complicated with it, cos(x)dx=d(sin(x)). Just substitute u=sin(x). There are rules for dealing with powers of sin's and cos's. They are particularly easy if one power is odd.
     
  6. Oct 19, 2009 #5
    Ok I know what your talking about. It was just in our chapter about integration by parts so I was a little first sight shocked. Can some one give me an answer so when I complete it I can know if I am correct or not.
     
  7. Oct 19, 2009 #6

    Dick

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    Oh, come on. Just work it out and show us what you get. I'll guarantee someone will check it.
     
  8. Oct 19, 2009 #7
    Ok I have other work to do this one has bee on my mind all night. Ill post it tomorrow when I am clear of thought
     
  9. Oct 19, 2009 #8

    Dick

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    It's REALLY easy with the substitution lanedance suggested. You might want to clear your mind on this one and go to bed happy. But tomorrow is ok too.
     
  10. Oct 19, 2009 #9
    Ok thanks alot Ill see what I can do tonight.
     
  11. Oct 19, 2009 #10

    Dick

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    Free hint since you are playing along: cos^2(x)=1-sin^2(x)=(1-u^2).
     
  12. Oct 19, 2009 #11
    ok i got
    41(sin^7(x)/7)(1/2 x + 1/4sin2x+c)
     
  13. Oct 20, 2009 #12

    lanedance

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    doesn;t look quite right to me, maybe show your working
     
  14. Oct 20, 2009 #13
    Thats what I was thinking. after using u substitution I was left with

    (u)^6 (cos^2(x)) cos(x) du/cos(x)

    So cos(x) canceled out.
     
    Last edited: Oct 20, 2009
  15. Oct 20, 2009 #14

    Dick

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    Fine. Now what's cos^2(x) in terms of u?
     
  16. Oct 20, 2009 #15
    Oh ok the identity. so now all turns into
    u^7 (1-u^2) du
    Do I do their antiderivative now? And then substitute sin(x) back in. Im not sure if you can do each of there AD since they are multiplying one another.
     
  17. Oct 20, 2009 #16

    Dick

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    Of course you don't take the AD of each one. That's wrong. You multiply it out.
     
  18. Oct 20, 2009 #17
    OK that what I thought Im trying to do this not with out witting it down since Im in chemistry.
    So we get (u^6)-(u^8)du
    then (u^7)/7) - (u^9)/9)
    41 (sin^7(x))/7) - (sin^9(x))/9)
    is that it maybe?
     
  19. Oct 20, 2009 #18

    Dick

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    You are missing a parenthesis level following the 41, but yes, that's it.
     
  20. Oct 20, 2009 #19
    Ok awesome thank you for you patience.
     
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