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Ultra basic - integration

  1. Mar 4, 2006 #1
    Could anyone please explain to me very simply step by step how to integrate sin 2x (I know the answer is -(1/2) cos 2x) but how do you get there? Thanks please stay simple :-)
     
  2. jcsd
  3. Mar 4, 2006 #2
    the best way to think about integration in general is the opposite of differentiation if you can visualise what would differentiate to what you are trying to integrate you have cracked it


    for this specific question though, cos differentiates to - sin...... so sin integrates to - cos

    when you differentiate cos you bring the constant of x to the front, so instead bring the constant to the front but divide 1 by it

    using both of these you get your answer, hope it helped, im not the best at explaining things
     
  4. Mar 4, 2006 #3

    StatusX

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    Well, it should be pretty clear just from knowing how to take derivatives of things like sin(2x) what function would give you sin(2x) when you take its derivative. But if you need a more mechanical approach, used substitution with u=2x.
     
  5. Mar 4, 2006 #4
    should I use the product rule? doesn't it work like sin2x = sin(u)

    Yes I would love a mechanical approach, I often only really understand it only like that... could anyone give me the formula and the steps... thanks
     
  6. Mar 4, 2006 #5
    it isnt a product so no the product rule should not be used, its just 1 basic rule

    "the integral of sin(ax) = -1/a cos(ax)"

    ive explained why (although not too well) in my post above
     
  7. Mar 4, 2006 #6

    Ok great! thanks just what I need to get it... so if I want to integrate
    cos 2x I get (1/2)sin 2x
    sin 3x would give -(1/3) cos 3x

    cos 3x would give (1/3) sin 3x

    is this correct?
     
  8. Mar 4, 2006 #7

    Ok great! thanks just what I need to get it... so if I want to integrate

    cos 2x I get (1/2)sin 2x

    sin 3x would give -(1/3) cos 3x

    cos 3x would give (1/3) sin 3x

    is this correct?
     
  9. Mar 4, 2006 #8
    exactly, now go have lots of fun integrating

    (god wish i wasn't this ill so i could go drinking, saturday nights in are boring lol)
     
  10. Mar 4, 2006 #9

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    The mechanical approach is:

    [tex] \int \sin(ax) dx [/tex]

    take u=ax, du=adx, so dx=du/a:

    [tex] =\int \sin(u) du/a = \frac{1}{a} (-\cos(u)) = \frac{-1}{a} \cos(ax)[/tex]
     
  11. Mar 4, 2006 #10
    That's brilliant thanks everyone :-)
     
  12. Mar 4, 2006 #11

    HallsofIvy

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    Notice that integration by substitution is the chain rule in reverse.
     
  13. Mar 4, 2006 #12
    Also, when you are working on integration it is nice to have something that you can check your work with. A TI-89 is helpful and maple and mathematica are very nice, but a free web based version exists:
    http://integrals.wolfram.com/index.jsp

    You can plug some really tough integrals in there and get the answer.
     
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