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Incomplet notes on integration by parts

  1. Mar 26, 2005 #1
    i'm trying to complete my notets from my calculus II class. my professor showed us how to do the following integral using integration by parts but i'm not following his reasoning could some one fill me in on what i'm missing. thx in advance.


    let [tex]\ u=3x[/tex] so that [tex]\ du=3dx[/tex]

    let [tex]\ dv= \sin\frac{x}{2}[/tex] so that [tex]\ v= -2\cos\frac{x}{2}[/tex]

    now [tex]\int^{\pi}_03x\sin\frac{x}{2}\\{dx} = -6x\cos\frac{x}{2} \right]]^{\pi}_0\\+ 12\sin\frac{x}{2}\right]]_0^{\pi}[/tex]

    what is the intermediate step that i'm missing?
  2. jcsd
  3. Mar 26, 2005 #2
    If the integand is of the form of udv then this is how integration by parts goes:

    [tex]\int u dv = uv - \int v du[/tex]

    Integration by parts is possible only when dv can be integrated to get v.
    For example in your problem, dv = sin x/2. integrating this gives v=-2 cos x/2.

    If you are interested in how the above formula is derived,

    start from d(uv)

    d(uv) = u dv + v du

    Integrating this in both sides, gives

    [tex]\int d(uv) = \int u dv - \int v du[/tex]

    [tex]uv = \int u dv - \int v du[/tex]

    rearrange to get

    [tex]\int u dv = uv - \int v du[/tex]
  4. Mar 26, 2005 #3
    The product rule for derivatives is:

    [tex] udv + vdu = duv [/tex]

    Integrating both sides gives:

    [tex] \int{u}{dv} + \int{v}{du} = uv [/tex]

    Rearranging by subtracting [tex] \int{v}{du} [/tex] from both sides:

    [tex] \int{u}{dv} = uv - \int{v}{du} [/tex]

    If you plug in your u's and v's, you'll get what he has.
  5. Mar 26, 2005 #4
    By the way, honestly, your picture scares me. Specially when everyone else is sleeping and I am all alone here.
  6. Mar 26, 2005 #5
    thank you. and er... i'll change the pic. lol
  7. Mar 26, 2005 #6


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    A little ammendment

    [tex] dv=\sin\frac{x}{2} \ dx [/tex] (!!)

  8. Mar 26, 2005 #7


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    You're missing the actual [tex] \int{u}{dv} = uv - \int{v}{du} [/tex] part.

    In this case, [tex] \int{v}{du} = \int{3 *-2 \cos \frac{x}{2}}{dx}=\int{-6 \cos \frac{x}{2}}{dx}[/tex]
  9. Mar 26, 2005 #8
    He has not missed it. He has this already integrated

    [tex] \int{v}{du} = \int{3 *-2 \cos \frac{x}{2}}{dx}=\int{-6 \cos \frac{x}{2}}{dx} = 12\sin\frac{x}{2}[/tex]
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