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Do you HAVE always to use u-substitution?

  1. Oct 27, 2013 #1
    Do you always have to use u-substitution? i.e. sometimes will you get a different answer to if you didn't? Im asking because I had the integral ∫sinxcosx dx between 0 and π/2 and the answer was 1/2 but I got 0.

    ∫sinxcosx dx between 0 and π/2=-cos(π/2)sin(π/2)=0 according to my calculations.

    What am I doing wrong?

    I don't thin this counts as homework style as I am hoping just to find out whether you always have to use u-substitution in certain case using the above question as proof.
     
    Last edited by a moderator: Oct 31, 2013
  2. jcsd
  3. Oct 27, 2013 #2

    arildno

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    Antidifferentiating incorrectly.

    You are using a fallacious integration rule here:
    [tex]\int{u}'(x)v'(x)dx=u(x)v(x)+C[/tex]
     
  4. Oct 27, 2013 #3
    why is that not valid?
    what should I have done?
     
  5. Oct 27, 2013 #4

    arildno

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    You might try to differentiate u(x)*v(x).
    Do you then get your integrand u'(x)*v'(x)?

    If you don't get that result, then you've proven that u(x)*v(x) is not an antiderivative of u'(x)*v'(x)
     
  6. Oct 27, 2013 #5

    SteamKing

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    If you look at the original integral carefully, you would see that it has the form:

    [itex]\int u\ du[/itex]

    and it should be pretty obvious how to handle it.
     
  7. Oct 27, 2013 #6

    jtbell

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    For further advice on this exercise, please use the homework forums.
     
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