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Integration by parts HELP

  1. Jun 14, 2005 #1
    Integration by parts HELP !!!!

    Ok to be honest with all of you reading this post, i just dont understand how integration by parts work.
    Can someone please explain how it works?
    I have looked on the internet for help reading through all the notes but i still do not understand.
    So please somone explain to me in the easiest way.

    Here is an example :

    Integrate (2x Sin 3x) dx

    :uhh: :uhh: :uhh: :uhh:
  2. jcsd
  3. Jun 14, 2005 #2


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    1. Do you understand the product rule of differentiation?
    Essentially, integration by parts is nothing but the reverse operation of the product rule.

    2. Let us construct a function h(x)=u(x)v(x)
    Then, the product rule gives us:
    This can of course be rewritten as:
    This is the form of the product rule we use in integration by parts.

    3. Integration by parts.
    Suppose we may write an integrand in the form of [tex]u(x)\frac{dv}{dx}[/tex]
    (in your particular example, we may choose, say [tex]u(x)=2x, \frac{dv}{dx}=\sin(3x)[/tex])
    Then, from 2, we have:

    Are you following thus far?
    Last edited: Jun 14, 2005
  4. Jun 14, 2005 #3
    hello there

    the only time you would use integration by parts is when you have the product of 2 functions one in which you can reduce which is the [tex]u(x)=2x[/tex] through differentiation so that it equals to a constant or you would want to reduce it to something that will get eliminated once multiplied by the integral of [tex]\frac{dv}{dx}[/tex] so that it gets reduced to something that is integrable for example try doing integration by parts for

    [tex]\int \log x dx[/tex]

    if you can do this example and the one you are having trouble with you would pretty much prepared to use integration by parts when necessary
    Last edited: Jun 14, 2005
  5. Jun 14, 2005 #4
    I see, so one wouldn't want to use integration by parts to integrate sin(x) * e^x, then ;)
  6. Jun 14, 2005 #5
    woops i forgot one more sanario thanxs muzza lucky you mentioned it, a person tryin to do that for the first time would probably end up in circles lol thats pretty complicated to explain unless you actually do it,
  7. Jun 14, 2005 #6
    That integral requires integration by parts twice and then some rearranging, but it can be solved by this method.
  8. Jun 14, 2005 #7
    As for the original integral....


    dv = sin(3x)
    u = 2x

    [tex]\int udv = uv - \int vdu[/tex]

    [tex]\int2x\sin{(3x)}dx = 2x\int sin{(3x)}dx - \int 6\cos{(3x)}dx[/tex]
  9. Jun 14, 2005 #8
    Yes, precisely my point...
  10. Jun 14, 2005 #9
    Hmmm ??

    Ok thanks for all your reply but for the problem i have, so far i understand the rule :

    integrate u dv = u v - integrate v du

    i have put all the bits into the equation from this problem and it looks something like this, i dont know if its right.

    integrate (2x)(Sin 3x dx) = 2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)

    now here is the problem what do i do after this ??????

    Please help

    Thanks a bunch !!!!!
  11. Jun 14, 2005 #10


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    Continue until you solve it (meaning you might have to do integration by parts or integration by substitution again and again and again and again until everything is killed off)

    In case you succeed make sure to check for correct answer:

    [tex]\frac{2}{9} sin(3x) - \frac{2}{3}x cos(3x) + C[/tex]
  12. Jun 14, 2005 #11
    hmm don't understand lol

    Ok i get what u mean by integrating the equation but, how do i do it :

    can u show me the steps because i mean there are two integration i have to do but how do i but them together ???

    please show the steps and explain if you can !!!!!
  13. Jun 14, 2005 #12
    hello tin0

    well all you have to do is evaluate the right hand side and this is done by integrating the second term, you do know how to integrate cos3x coz really thats all you have to get your answer, you have already integrated sin3x so you should be able to do this integration, anyway i hope i have given you more than enough hints to get you answer
  14. Jun 14, 2005 #13


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    As you've said it yourself your problem now simplified to
    2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)

    Now go ahead and find [tex]\int{\frac{-2}{3} cos(3x) dx[/tex] using substitution method (try substituting something in cosine to make it a very easy trig function that can be easily integrated, and dont forget to check for any additional things you would need after substituion (like perhaps checking if your new substituted's variable derivative has enough numbers for that original integral)
  15. Jun 15, 2005 #14
    I tried

    So far i got to is :

    integrate (2x Sin 3x) dx

    u = 2x
    dv = Sin 3x dx
    du/dx = 2
    du = 2 dx
    v = -1/3 Cos 3x

    integrate (2x)(Sin 3x dx) = 2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)
    = 2x (-1/3 Cos 3x) - integrate -2/3 Cos (3x) dx

    But i have never even learnt how to use substitution by trigometry method so is there any other way i can do it ???

    if i do integration by substitution with this :

    integrate -2/3 Cos (3x) dx

    u = -2/3 Cos (3x) or u = -2/3 Cos

    du/dx = 2/3 3Sin 3x or du/dx = 2/3 Sin

    I dont know what to do this is really confusing lol ?????
  16. Jun 15, 2005 #15
    hello there

    well lets use another dummy variable r since we have already used u anyway
    let r=3x
    dx=dr/3 then use the substitution method .
    now the most easiest way to do it is to look up a full table of standard integals and look for something that looks like what you are working with, it isnt that complicated

  17. Jun 15, 2005 #16
    I am still confused !!!

    Ok integration by substitution :

    integrate -2/3 Cos (3x) dx

    r = -2/3 cos 3x why r = 3x ??????????????


    can u show me the steps how to do it ????
  18. Jun 15, 2005 #17


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    You can take out the coefficients outside because they are not to be integrated - they are just coefficients without the 'x' so to speak. The simplest trig function for you to do is cos(u). So you have 2/3 integral ( cos(3x) dx). Now what do you think you should do to make it into 2/3 integral (cos(u) du)? You substitute:

    u = 3x
    du = 3dx

    1/3 * -2/3 integrate (cos(u) du)

    Since you need 3dx in order to substitute for u you have to multiply 3 by cos(u), but now you must balance it out by multiplying by 1/3 outside. Hence your answer ends up being 2/9 sin(3x) for this integral. Add it up with previous results and your answer is

    2/9 sin(3x) - 2/3 x*cos(3x) + C
    Last edited: Jun 15, 2005
  19. Jun 15, 2005 #18

    Thanks alot people i actucally sort of understand now, thanks alot !!!!!!

    Thanks peeps !!!!!
  20. Jun 15, 2005 #19
    One more little question !!!

    The question was:
    integrate (2x Sin 3x) dx

    and i put this into :

    integrate (2x)(Sin 3x dx) = 2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)

    i have asked my teacher and he told me to leave 2x (-1/3 Cos 3x) and not change it would that be right ?? or do i change it to :

    2/9 sin (3x)

  21. Jun 15, 2005 #20
    hello there

    this is the answer
    [tex]\frac{2}{9} sin(3x) - \frac{2}{3}x cos(3x) + C[/tex]
    there is nothing to change, listen to your teacher, especially if your teacher told you not to change anything why do you want to change it? if you want to change the order of the terms around in the answer feel free to do so because that wont change the answer anyway good luck with it

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