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Integration help

  1. Mar 5, 2008 #1
    I'm currently learning integration by trig substitution. The problem is:
    integral ( x^2/sqrt(16-x^2) ) dx.
    I seem to be having some difficulty...this is what I have:
    let x=4sinu
    dx = 4cosu du

    Thus, we have:

    integral 16sin^2(u)*4cosu du/(4cos^2(u)) (the last part is from the trig identity sin^2x + cos^2x = 1)

    But now what? I simplied that as much as I could and then used integration by parts, but I ended up with the wrong answer. The correct answer is 8arcsin(x/4) - 0.5x*sqrt(16-x^2). Have I made a mistake in what I've done so far? If not, where do I proceed?

  2. jcsd
  3. Mar 5, 2008 #2
    [tex]\sqrt{1-\sin^2u}=|\cos u|[/tex].
  4. Mar 6, 2008 #3

    Gib Z

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    Check your denominator again. Should the cosine term really be squared? Shouldn't be any need for integration by parts after that.
  5. Mar 6, 2008 #4
    [tex]\int \frac{x^2}{\sqrt{16-x^2}}dx[/tex]

    [tex]x=4sin u, dx=4cos u du[/tex]

    [tex]\int\frac{16sin^2u}{(4cosu)}4cosu du[/tex]

    [tex]16\int sin^2udu[/tex]






    Your mistake was that your denominator should have been 4cosu, not [tex]4cos^2u[/tex]
  6. Mar 7, 2008 #5

    Gib Z

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    Your new so I won't say much, but please 1) Do not give out a full solution, it never helps the original poster learn anything and 2) Repeat what another member says just in a more blatantly obvious manner.
  7. Mar 7, 2008 #6
    Sorry and thanks
  8. Mar 7, 2008 #7

    Gib Z

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    I feel bad for being so harsh now :( Sorry lol.
  9. Mar 7, 2008 #8


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    Gold Member
    Dearly Missed

    I don't.
    I'll flay him him next time.

    Flebbyman: I know your adress..:devil:
  10. Mar 7, 2008 #9
    Don't sweat it

    :surprised :smile:
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