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Homework Help: Integration problem using substitution

  1. Feb 1, 2017 #1

    chwala

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    1. The problem statement, all variables and given/known data

    using ## u= sin 4x## find the exact value of ##∫ (cos^3 4x) dx##



    2. Relevant equations


    3. The attempt at a solution

    ## u= sin 4x##
    on integration ##u^2/2=-cos4x/4 ## , →##-2u^6={cos 4x}^3 ##......am i on the right track because now i end up with ##∫{{-2u^6}/{4.-2u^2}}du## or should i use

    ##du=4cos 4x dx ## to end up with ## 0.25 ∫ cos^24x du## which looks wrong to me
     
  2. jcsd
  3. Feb 1, 2017 #2

    haruspex

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    No, you've integrated one side wrt u and the other wrt x.
     
  4. Feb 1, 2017 #3

    chwala

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    so should i use the second approach?
     
  5. Feb 1, 2017 #4

    haruspex

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    Yes, but you need to get all of the references to x turned into references to u.

    You are asked for an exact value, but it is an indefinite integral. Remember that the limits need to be expressed in terms of u as well.
     
  6. Feb 1, 2017 #5

    chwala

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    yes the limits are from 0 to π/24
     
  7. Feb 1, 2017 #6

    chwala

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    is ## 0.25∫{cos^24x}du## correct?
     
  8. Feb 1, 2017 #7

    haruspex

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    So what are the limits on u?
    Yes.
     
  9. Feb 1, 2017 #8

    chwala

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    limits on u are 0 to 30, now how do i proceed with the integration?
     
  10. Feb 1, 2017 #9

    chwala

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    ## cos^2 4x## = ##(cos 8x+1)##/2

    should we substitute again? or are we going to have


    ##0.25∫cos^24x d {sin4x} ##
     
  11. Feb 1, 2017 #10

    chwala

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    i am a bit confused we cannot integrate a variable say ##x## with respect to another variable say ##u##, i am stuck here
     
  12. Feb 1, 2017 #11

    haruspex

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    No.
    You have the cos2 of some angle, and you need to express that in terms of u, the sine of the same angle. Does nothing click?
     
  13. Feb 1, 2017 #12

    chwala

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    sorry limits are from 0 to 0.5 an oversight on my part..........
     
  14. Feb 1, 2017 #13

    chwala

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    i now get it lol
    ## 0.25∫{1-u^2}du ## from u=0 to u=0.5 thanks mate solution is ## 0.115##
     
  15. Feb 1, 2017 #14
    Why don't you try splitting ##cos^34x## into ##cos4x## and another term containing the term used for ##u## substitution?
     
  16. Feb 1, 2017 #15

    chwala

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    i have seen the obstacle with that move.........
     
  17. Feb 1, 2017 #16

    chwala

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    i have seen it, check post 13
     
  18. Feb 2, 2017 #17

    SammyS

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    @chwala
    Actually that is the move you finally made to solve. Check the time of @Eclair_de_XII 's post and your posts.
     
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