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What is the best way to integrate this?

  1. Feb 10, 2007 #1
    What is the best way to integrate this??

    What is the best way to integrate -250*cost*sin^2t dt ??

    I'm looking in my old calc book and it says to integrate by parts-is this the only way to do it? When I try this it makes the integral even more complicated than it was to begin with.
  2. jcsd
  3. Feb 10, 2007 #2


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    What are you choosing as u and dv?
  4. Feb 10, 2007 #3
    Try a u-substitution. The most obvious being u=sin(x)
  5. Feb 10, 2007 #4
    Okay so heres how I did it:

    The general rule is Integral of udv = u*v - int(v*du)

    So i said that v = sin^2t dv/dt = 2sintcost
    du=costdt --> u=sint

    Integral = sin^3t - int (sin^2t*costdt)

  6. Feb 10, 2007 #5
    But then what do you do with cost ?? how do you put that in terms of u??
  7. Feb 10, 2007 #6

    This is really messy, are you absolutely sure the book said to use integration by parts because if the problem is exactly as you've posted it then there is a much simpler way.
  8. Feb 10, 2007 #7
    Well how do you normally do a u-substitution? What is du?
  9. Feb 10, 2007 #8
    No it absolutely didnt say intgrate by parts...i'm in a higher level class and havent touched calc in 5 years and so i forget how to do this basic operation. I looked in my old freshman yr calc book and thats what it said to do. If you have a simpler way, i'm sure thats what we should do.
  10. Feb 10, 2007 #9
    We have to do line integrals and as a part of the g(x,y)dx, dy, ds..this was one of the problems i was having (actually performing the integral).
  11. Feb 10, 2007 #10
    I don't even believe a freshman level book would say to do integration by parts for this problem.
  12. Feb 10, 2007 #11
    um. okay. well i'm not lying?? I mean i dont know how that it supposed to be constructive. do you have the name of the method, the easier way, that you would use?
  13. Feb 10, 2007 #12
    Yes I do, and I've mentioned it twice already.
  14. Feb 10, 2007 #13
    Ooooh. got it, thanks. so when should i use the u substitution method vs integration by parts? the u sub worked out perfectly in this case...
  15. Feb 10, 2007 #14
    Parts often seems to be reserved for a last resort situation when nothing else seems to work, so it's probably best to try and use a u-substitution or other method before resorting to parts.
  16. Feb 10, 2007 #15


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    Another way to do it would be to use double angle formula (for cos) and then factor formulae to simplify, then just integrate. Not difficult either.
  17. Feb 11, 2007 #16
    You could use the double angle of sin. (sin(2t)=2sintcost) and substitut it .

    Would be somethink like this: -250 S cos t sin 2t dt
    -250 S cos t 2sint cost dt
    -500 S (cos t)^2 sint dt
    Then u=cost du=-sint dt, then 500 S u^2 dt
    500 (u^3)/3
    substituting back u= cost 500 (cost)^3* 1 / 3

    Wala, there is, no integration by parts is useless if you have the du on the integral. Just the double angle formula.

  18. Feb 11, 2007 #17
    With all respect,
    You are the rule right but the dv/dt= -2 cos 2t, you just use the double angle formula.

  19. Feb 12, 2007 #18


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    The original problem does not contain sin(2t), it has sin^2t.
    As d_leet has mentioned, it appears, that the simplest way is to put u=sint.
    Last edited: Feb 12, 2007
  20. Feb 14, 2007 #19
    OHHH, I didn't notice, the that's the way, you are right.
  21. Feb 14, 2007 #20

    D H

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    This one is quite easy with integration by parts. It's no harder (and maybe even easier) than u-substitution.

    [tex]\int \sin^2t \cos t \; dt[/tex]

    Applying integration by parts with
    [tex]u = \sin^2 t, dv = \cos t\;dt \text{\ for which\ } du = 2 \sin t \cos t\; dt, v = \sin t[/tex]
    [tex]\int \sin^2t \cos t \; dt = \sin^3 t - 2 \int \sin^2t \cos t \; dt[/tex]
    The integral on the left- and right-hand sides are the same integral. Thus
    [tex]3 \int \sin^2t \cos t \; dt = \sin^3 t [/tex]
    [tex]\int \sin^2t \cos t \; dt = \frac{\sin^3 t}{3}[/tex]
    Last edited: Feb 14, 2007
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