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Help with Integration

  1. Feb 4, 2006 #1
    How, or where can I find a solution to Integrals of :

    |cos θ| * ½ (sin θ)^2

    AND
    |cos θ| * ½ (1 + sin θ)^2

    to solve from 0 to 2π

    I think the first should be the same as:
    cos θ * ½ (sin θ)2 :
    Integrated from 0 to π/2, times 4

    Note : “| |” is for absolute value
    RB
     
  2. jcsd
  3. Feb 4, 2006 #2

    In either case you can make a u substitution of u = sinθ, but also in both cases you will have to split the integral into a few integrals because of the absolute value.
     
  4. Feb 5, 2006 #3

    VietDao29

    User Avatar
    Homework Helper

    Yes, this is correct. Do you know how to show it?
    Or you can do it normally by splitting the integral into smaller ones, and try to get rid of the absolute value. I'll give you a simple example:
    [tex]\int \limits_{-2} ^ 1 |x| dx[/tex]
    Since we have |x| = x, for x >= 0; and |x| = -x, for x < 0, we can do as follow:
    [tex]\int \limits_{-2} ^ {1} |x| dx = \int \limits_{-2} ^ 0 |x| dx + \int \limits_{0} ^ {1} |x| dx = \int \limits_{-2} ^ 0 (-x) dx + \int \limits_{0} ^ {1} x dx[/tex]
    [tex]\left. -\frac{x ^ 2}{2} \right|_{-2} ^ 0 + \left. \frac{x ^ 2}{2} \right|_{0} ^ 1 = 0 - \left( -\frac{4}{2} \right) + \frac{1}{2} - 0 = \frac{5}{2}[/tex]
    Now we have |cos (x)| = cos(x), if [tex]x \in \left[ -\frac{\pi}{2} + 2k \pi, \frac{\pi}{2} + 2k \pi \right], \quad k \in \mathbb{Z}[/tex], and
    |cos (x)| = -cos(x), if [tex]x \in \left] \frac{\pi}{2} + 2k \pi, \frac{3 \pi}{2} + 2k \pi \right[, \quad k \in \mathbb{Z}[/tex]
    Can you do those integrals now? :)
     
  5. Feb 5, 2006 #4
    Actually no,
    I’m not looking to DO an Integral,
    Or find a solution to HOW to do an Integral, (This isn’t homework or some test).
    I just need the solution for area from angle 0 degrees to 360 degrees. (The “PI” didn’t display very well in the first OP for “0 to 2 pi” and “0 to pi/2”.

    Example I think I can pull the ½ factor outside the integration - but I don’t 100% know, Nor do I know how to work with a cos multiplied by a sin squared inside (or the addition inside the sq.) all inside the integration.

    I just need the solution to total surface area under these two curves to the axis (i.e. Adding not subtracting the area under axis line)
    I don’t need to learn how to do it myself for these two answers.

    If the answers are easy great, If it’s tricky suggestions on where to go for help is fine as well.
    Thanks RB
     
  6. Feb 6, 2006 #5

    VietDao29

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    Homework Helper

    Okay, this is fine.
    [tex]\int \limits_{0} ^ {2 \pi} \frac{1}{2} |\cos x| \sin ^ 2 x dx = \frac{2}{3}[/tex]
    [tex]\int \limits_{0} ^ {2 \pi} \frac{1}{2} |\cos x| (1 + \sin x) ^ 2 dx = \frac{8}{3}[/tex]
    [tex]\int \limits_{0} ^ {\frac{\pi}{2}} \frac{1}{2} |\cos x| \sin ^ 2 x dx = \frac{1}{6}[/tex]
    [tex]\int \limits_{0} ^ {\frac{\pi}{2}} \frac{1}{2} |\cos x| (1 + \sin x) ^ 2 dx = \frac{7}{6}[/tex]
    Is this what you need? :)
     
  7. Feb 6, 2006 #6
    Excellent
    thanks
    RB
     
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