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Intigration help

  1. Aug 2, 2007 #1
    The question is Intigrate [tex]\sqrt{1-Sin2x}[/tex]

    this was my attempt to solve the question

    sin2x = 2sinxcosx

    (1-2sinxcosx)^1/2

    u = (1-2sinxcosx)


    [tex]\frac{du}{dx}[/tex] = (x-2cos[tex]^{}2[/tex]x)

    du = (x-2cos[tex]^{}2[/tex]x) dx

    im not completely confident about the u substitution rule becouse i have not learned it yet at school. this is as far as i can go with reading in my textbook. is this the correct method to approch this type of question? and if it is what should i do from here?
     
  2. jcsd
  3. Aug 2, 2007 #2

    Gib Z

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    [tex] \int \sqrt{ 1 - \sin (2x)} dx = \int \sqrt{ 1 - 2\sin x \cos x} dx[/tex]
    [tex]= \int \sqrt{ \sin^2 x - 2\sin x \cos x + \cos^2 x} dx[/tex]
    [tex] = \int \sqrt{ ( \sin x - \cos x)^2} dx = \int \sin x - \cos x dx[/tex]
    [tex] = -( \sin x + \cos x) + C[/tex]
     
  4. Aug 2, 2007 #3
    Thanks GibZ, that's a good one.
     
  5. Aug 2, 2007 #4
    wow that worksout so nicely thnx alot!
     
  6. Aug 2, 2007 #5

    Gib Z

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    Quite Proud I made the answer turn out so nicely actually :) The Integrator gives some ugly thing.
     
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