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Integral Question

  1. Apr 5, 2009 #1
    I am new to this forum, please bear with me.

    1. The problem statement, all variables and given/known data

    dy/dx=xcos^2(x)



    3. The attempt at a solution

    I've already gotten cos^2(x)=1/2(cos2X) but I don't know what to do with that.

    Thanks for your help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 5, 2009 #2

    Cyosis

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    Homework Helper

    You made a mistake applying the double angle formula, it should be [itex]\cos^2 x=\frac{1}{2}(\cos 2x + 1)[/itex]. Now use integration by parts to get rid of the x.
     
    Last edited: Apr 5, 2009
  4. Apr 5, 2009 #3
    Yes, I typed in the D.A. formula wrong. I had it right on my paper though. We have not gotten to integration by parts yet. Is there another way of doing this with a u-substitution?

    Thanks again for your help.
     
  5. Apr 5, 2009 #4
    Nope, you need integration by parts here.
     
  6. Apr 5, 2009 #5
    If you haven't gotten to integration by parts yet, then I would double check that you wrote the problem down correctly. If you are just learning u-substitution, then I think the problem may be
    [tex] \int x\cos (x^2) \,dx [/tex]

    The problem that you posted is a typical integration by parts problem.
     
  7. Apr 5, 2009 #6
    Thank You everyone. I feel like I have wasted your valuable time. I did indeed misread the problem. You are correct n!kofeyn, that is exactly what the problem is asking for.

    Again, sorry for bothering you all.
     
  8. Apr 5, 2009 #7
    You have not wasted our time, nor have you bothered us. We wouldn't bother posting if we felt you were.

    There is often confusion on this notation, as it can be unclear whether
    [tex] \cos x^2 = \cos(x^2)[/tex] or [tex] \cos x^2 = (\cos x)^2 = \cos^2x [/tex]
    (The first is usually what is meant.)

    Please continue to post if you have questions!
     
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