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The integral of (sin 3t)^5 cos t dt

  1. Mar 27, 2005 #1
    Hey folks,

    HERE GOES.... :confused:

    the integral of (sin 3t)^5 cos t dt

    i believe you have to use u substitution but i am having trouble getting it set up correctly.

    Thanks for any input

    :smile:
     
  2. jcsd
  3. Mar 27, 2005 #2
    [tex] \int{sin(3t)^5}cos(t)}{dt} [/tex]

    u = sin(3t)
    du = 3cos(3t) dt

    [tex] 1/3\int{u^5*3cos(3t)}{dt} [/tex]

    [tex] 1/3\int{u^5}{du} [/tex]
     
  4. Mar 27, 2005 #3
    How does that work? What happened to the cos t?
     
  5. Mar 27, 2005 #4
    From converting the integral to one with respect to u instead of t. The conversion factor is right above the integral
     
  6. Mar 27, 2005 #5
    Ugh I did it wrong hold on. Im so tired
     
  7. Mar 27, 2005 #6

    Galileo

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    There may be easier ways, but you could expand:

    [tex]\sin(3t)=4sin(t)cos(t)^2-sin(t)[/tex]
    by exploiting deMoivre's theorem.
     
  8. Mar 27, 2005 #7
    Mathworld's Integrator gives a long and not-very-nice answer btw.
     
  9. Mar 27, 2005 #8
    Then you'd have to raise that to the fifth and foil and blah blah blah. Its a real complicated integral.
     
  10. Mar 27, 2005 #9

    Galileo

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    Well, you ARE multiplying a power of a sin with a cosine with a different argument, I don't expect the answer to be nice.
    Besides, you just have to develop a 'work-attitude' in some situations. Roll up your sleeves and do it. It may be the fastest way. It won't take more than a few minutes, while looking for a possible easier way probably takes longer.
     
  11. Mar 27, 2005 #10
    [tex] sin(3t) = sin(2t+t) = sin(2t)cos(t) + sin(t)cos(2t) = 2sin(t)cos^2(t) + sin(t)(cos^2(t)-sin^2(t)) [/tex]

    [tex] = 2sin(t)(1-sin^2(t)) + sin(t)(1-2sin^2(t)) = 2sin(t)-2sin^3(t) + sin(t)-2sin^3(t) = 3sin(t)-4sin^3(t)[/tex]



    [tex] \int{sin(3t)^5cos(t)}{dt} [/tex]

    [tex] \int{(3sin(t)-5sin^3(t))^5cos(t)}{dt} [/tex]

    [tex] u=sin(t), du = cos(t) dt[/tex]

    [tex] \int{(3u-4u^3)^5}{du} [/tex]

    I think thats as good as it gets if I didnt make any errors.
     
    Last edited: Mar 27, 2005
  12. Mar 27, 2005 #11

    dextercioby

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    This is integral is the typical example of an easy & but messy integral.Eas,because you know what to do to get to the result and messy,because it would take a page of writing to do it...

    Daniel.

    P.S.[tex] \sin 3x=-4\sin^{3}x+3\sin x [/tex]
     
  13. Mar 27, 2005 #12

    dextercioby

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    [tex] \int \left( \left( \sin 3x\right) ^5\cos x\right) dx= -\frac 1{512}\cos 16x-\frac 1{448}\cos 14x+\frac 1{64}\cos 10x+ \frac 5{256}\cos 8x-\frac 5{64}\cos 4x-\frac 5{32}\cos 2x +C [/tex]

    Daniel.
     
  14. Mar 27, 2005 #13

    dextercioby

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    Making the substitution

    [tex] \sin x=u [/tex]

    u'll need to evaluate this cutie pie

    [tex] \int \left(-4u^3+3u\right)^{5} \ du [/tex]

    So use the binomial formula.

    Daniel.
     
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