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Integration question

  1. Jan 28, 2006 #1
    I have been asked to find the integral sinx cox dx using the identity sin2x = 2sinxcosx

    My work...

    integral of sinx cox dx

    = 1/2 integral of 2 sinx cos dx

    = 1/2 integral of sin 2x dx

    u = 2x
    du = 2

    so 1/2 * 1/2 of integral of sin u du

    = 1/4 [-cos u] + c
    = - 1/4 cos 2x + c is this correct?
     
  2. jcsd
  3. Jan 28, 2006 #2

    arildno

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    Quite so! :smile:
     
  4. Jan 28, 2006 #3

    0rthodontist

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    Yes, it is, except you should say du = 2 dx instead of just 2.
     
  5. Jan 28, 2006 #4
    Great. So if we know look at the following integral x sin x cos x dx

    My work....

    Let u = x

    du/dx = 1

    dv/dx = sin x cos x

    v = -1/4 cos 2x (from above)

    so = -1/4 x cos x - integral of -1/4 cos 2x dx

    = -1/4 x cos 2x + 1/4 integral of cos 2x dx

    Let u = 2x

    du/dx = 2

    so = -1/4 x cos 2x + 1/4 * 1/2 integral of cos u du

    = -1/4 x cos 2x + 1/8 sin 2x + c is this correct?
     
  6. Jan 28, 2006 #5

    arildno

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    Yes it is, as can be verified by differentiating your expression.
     
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