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Integration continued

  1. Jan 25, 2006 #1
    I have been asked to find the integral sinx cox dx and the integral x sinx cosx dx using the identity sin2x = 2sinxcosx

    I don't know what to do. Can anyone help please, hints,.... answers? :-)
     
  2. jcsd
  3. Jan 25, 2006 #2

    StatusX

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    You can integrate sin(2x), right?
     
  4. Jan 25, 2006 #3
    is it -2 cos x?
     
  5. Jan 25, 2006 #4

    StatusX

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    Don't guess. You can check to see if the derivative of that gives you back sin(2x). It doesn't, and doing that should help you see what would.
     
  6. Jan 25, 2006 #5

    VietDao29

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    As StatusX has pointed out, it's not!
    Looking at your integral table (if integration is new to you, then it's best to have an integral table with you when integrating), you should see something that reads:
    [tex]\int \sin x dx = - \cos x + C[/tex]
    Since x is a dummy variable, x can be anything, such as:
    [tex]\int \sin u du = - \cos u + C[/tex]
    [tex]\int \sin \left( e ^ x \right) d \left( e ^ x \right) = - \cos \left( e ^ x \right) + C[/tex]...
    So you should use a u-substitution here. What is u? (u = ?)
     
    Last edited: Jan 25, 2006
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