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Multiple trig integral

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Integrate: (sin(2x))^3(cos2x)^2dx

    2. Relevant equations

    Using substitution

    Cos2x= (1-(sinx)^2)

    3. The attempt at a solution

    I sub u= sin2x

    But then got nowhere because I had cos2x to the power of 2 and I don't know how to compensate for it with du.
     
  2. jcsd
  3. Sep 26, 2014 #2

    haruspex

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    First, an easy substitution gets rid of the 2 factors in the 2x terms.
    When you have a mix of sin and cos in an integral dx, look for combining one of them with the dx, e.g. cos(x)dx = d sin(x).
    In the present case, you can choose a cos or a sin. Which works better?
     
  4. Sep 26, 2014 #3

    HallsofIvy

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    When you have even powers of both sine and cosine, reduce the powers using [itex]sin^2(x)= (1/2)(1- cos(2x))[/itex] and [itex]cos^2(x)= (1/2)(1+ cos(2x))[/itex].
     
  5. Sep 26, 2014 #4

    pasmith

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    I can't tell whether you mean [itex]\cos^2 x= 1 - \sin^2 x[/itex], which is true, or [itex]\cos 2x = 1 - \sin^2 x[/itex], which is false: [itex]\cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x[/itex].

    You have an extra power of [itex]\sin 2x = -\frac12 \frac{d}{dx} \cos 2x[/itex]. That suggests [itex]u = \cos 2x[/itex], not [itex]u = \sin 2x[/itex].
     
  6. Sep 26, 2014 #5

    haruspex

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    It's simpler than that. See my post #2.
     
  7. Sep 26, 2014 #6
    I got it. You use he identity and make sin into 1-cos. and then sub u for cos 2x and work from there.
     
  8. Sep 28, 2014 #7

    HallsofIvy

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    Please, please, please be more careful about what you are writing. You cannot "make sin into 1- cos"! They are not equal.
     
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