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Homework Help: Integral question

  1. Dec 15, 2007 #1
    For the integral

    cos^(5)(4x), can I just integrate it directly to -1/4*sin^(6)(4x)... or which technique should I use?
  2. jcsd
  3. Dec 15, 2007 #2
    Nope you can't do that. To make things easier, you may want to sub u=4x first, and then you have to use a trig identity to make the (cos(u))^5 "easy" to integrate.

    Hint: [tex]\sin^2{x} + \cos^2{x} = 1[/tex]
  4. Dec 15, 2007 #3
    what is the derivative of your answer then? it will not give you your original problem.

    remember, when you're dealing with powers, you have to take into consideration the "chain rule."
  5. Dec 15, 2007 #4
    But then I would have two 1-sin^2(X) in the integral, is there an easier way?
  6. Dec 15, 2007 #5
    As neutrino suggested, after obtaining cos^5(u) write it as cos^4(u)*cos(u) = ((1-sin^2(u))^2)*cos(u). Now giving sin(u)=v, can you proceed?
  7. Dec 15, 2007 #6
    Ok, so i get:

    (1-v^2)^2 dv

    = 1 - 2(v^2) + v^4...
  8. Dec 15, 2007 #7
    Now, I have to integrate the whole thing??????????????????????????????

    How would I sub back? --let v = sin(4x)???
  9. Dec 15, 2007 #8


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    v = sin(u) and u=4x.

    You've expanded the brackets in the post above but did you integrate it?
  10. Dec 15, 2007 #9
    Too much work for this question.
  11. Dec 15, 2007 #10


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    If you are only interested in questions that can be done trivially, then I recommend you drop calculus.
    Do you really consider it too difficult to integrate
    [tex]\frac{1}{4}\int (v^4- 2v^2+ 1)dv[/tex]?
  12. Dec 15, 2007 #11
    I second Halls of Ivy. I am not sure what is so difficult here. It is a matter of individually integrating 3 simple terms and then plugging in for v.

    In the time it took to write too much work for this question, you could have finished!

  13. Dec 15, 2007 #12
    In context, my statement is valid...
  14. Dec 15, 2007 #13
    I have no idea what this means.
    Last edited: Dec 15, 2007
  15. Dec 15, 2007 #14


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    I think frasifrasi means that they only wanted to know how to substitute back for v once the integration was done so it was too much work to put the full integration in the previous post for the relatively simple clarification.
  16. Dec 15, 2007 #15
    Yeah, I was just saying that more work wasn't really necessary, not meaning to complain in any way.

    but hey, I have another question...

    For the series (-1)^(n)/ln(n), I used the alternating series test to show that it converges, but how do I show that it converges conditionally?

    If I look at the abs value 1/ln(x) , I am not sure how to establish that it diverges...

    Thank you for the help!
    Last edited: Dec 15, 2007
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