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Integration by Parts guidelines

  1. Dec 2, 2009 #1
    I've been trying to find this online, but I haven't been able to find any site that really explains it: when performing integration by parts, is there some rule or set of guidelines to determine which part of the equation is u and which is dv?
     
  2. jcsd
  3. Dec 2, 2009 #2
    Check out PatrickJMT on youtube, he explains it in a couple of videos.
     
  4. Dec 3, 2009 #3

    Pengwuino

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    Gold Member

    For the most part, you want identify which part of integration you want to pull the derivative off of. For example if you have something like integrating [tex]Ln[x]x^2[/tex], it'd be very convenient to pull off a derivative from [tex]x^2[/tex] and put it on the natural log to turn it into a [tex]\frac{1}{x}[/tex] which makes the integration much easier.
     
  5. Feb 19, 2010 #4
    my calculus teacher said save the easiest part of the integral that you can still integrate for "dv" and use the rest for the "u". my textbook has more about that, and I'll get back to you later on it.
     
  6. Feb 19, 2010 #5
    There are a couple of ways to look at it. If you want a general strategy, try LIATE.

    Logarithms
    Inverse trig
    Algebraic
    Trigonometric
    Exponential.

    Let u=whichever expression comes first in your integrand; then let dv=the next.

    Ex:

    [tex]

    \int(2x\*\sin(x))dx

    [/tex]

    let [tex]u=2x[/tex] (because it is algebraic) and [tex]dv=\sin(x)dx[/tex] (because it is trigonometric)

    LIATE won't always work; it's just a nice first try type of strategy. However, I prefer to think of it this way:

    I choose the easiest function to integrate to be dv, and everything else to be differentiated because differentiation is easy and integration is harder. I can differentiate anything, but I can't integrate everything. In the above example, since neither expression is really any harder or easier to integrate, I might refer to the LIATE strategy to pick my subs. But it is good to be mindful of your substitutions and think ahead to what will happen.

    What would have happened if I'd let [tex]u=\sin(x)[/tex] and [tex]dv=2x dx[/tex] instead? I'd get a recursion that is not helpful: [tex]2x[/tex] becomes [tex]x^2[/tex], [tex]\sin(x)[/tex] becomes [tex]-\cos(x)[/tex] and I'm in no better position. Integrating by parts again, [tex]x^2[/tex] becomes [tex]x^3/3[/tex], and [tex]-\cos(x)[/tex] becomes [tex]-\sin(x)[/tex], which is again, not helpful.

    So try to think ahead, and see if you'll need to integrate by parts multiple times. If so, let u=whatever is going to "ratchet down" to zero-power by repeated differentiation.

    Hope this helps, cheers!
     
  7. Feb 19, 2010 #6
    alright check this out. the essence of integration by parts is that you are simplifying the integral so you can actually do it. For example, you will encounter integration by parts a lot with e^x. Basically you dont want the derivative of e^x to be in your final integral , because it doesn't simplify it at all. if you have xe^x, you will want the derivative of x to be in the final integral, because then you can easily do the integral of 1 * e^x dx, which is e^x + C
     
  8. Mar 25, 2010 #7
    My calc teacher taught us LIPET. Log, inverse trig, polynomial, exponential, trig. She says it always works and I've never found a problem where it doesn't. You pick u in the order of LIPET, and whatever remains is dv.
     
  9. May 26, 2010 #8
    I was taught to perform integration by parts using ILATE(or LIATE, depending upon the question).
    But as 2h2o rightly pointed out, you must always choose the terms you can easily integrate to be u.

    Now consider the example 2xsinx. If you choose to integrate 2x, then you will get x^2. By doing so, you will be unnecessarily increasing the power of x, when you could have simply selected sinx as the integrand. Thus, you must always try NOT to complicate the given function by increasing its powers.

    Hope it helps.
     
  10. May 26, 2010 #9
    In an exam, if you forget and you really don't know what to do for some reason, just do both and one of them will be obviously worse. (the integration and differentiation of the "parts" is usually pretty easy anyway)
     
  11. Jun 27, 2010 #10
    In general, this is how I was introduced to integration by parts. The first thing you look for is a product in your integrand. If you have a product, and simple u substitution does not work, then integration by parts is a viable option. Here is the general method I follow:

    ∫udv=uv-∫vdu


    Steps
    1. Choose u and dv.
    In general, you want to choose dv such that its antiderivative is simple (v).
    Additionally, you need to choose u such that its derivative is fairly simple. The idea is that you want the resulting integral (∫vdu) to be easier to integrate than the original integral.
    2. Find du (the derivative of u) and v (the antiderivative of dv). You now have all the terms you need in order to complete the above expression.



    NOTE: the idea of integration by parts is that you want to end up with an easier integral to integrate. If you end up with a more complicated integral, then maybe you have chosen the wrong u and dv. Alternatively, perhaps integration by parts is not the best method for said question.

    ex: ∫xexdx
    1. u= x, dv=ex
    2. du=dx (derivative of x is 1), v=ex (antiderivative of ex is ex)
    3. Our new expression is:

    I= ∫xexdx= xex-∫exdx

    Note that our new integral is much simpler and we immediately know its solution.

    our solution is then:
    I= xex-ex+C

    I Hope this Helps!
    -Curtis
     
  12. Jul 5, 2010 #11

    Redbelly98

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  13. Jul 5, 2010 #12
    Sometimes when it looks like IBP isn't "working", you might need to do a little trick. Look up the integral of cosine times exp.
     
  14. May 9, 2011 #13
    it is not integration by parts it is by substitution
    for example,
    integral of sin^3x / cosx dx=integral of( sinx)(sin^2(x) / cosx dx and then you set : t=cosx==>
    dt=(cos(x))'dx=-sin(x)dx and sin^2(x)=1-cos^2(x)(since sin^2x+cos^2x=1 Pythag.identity)
    Then the integral becomes:integral of-(1-t^2)/tdt=integralof(-1/t-t^2/t)dt(… divided the numerator by dinominator)=integral of(-1/t+t)dt=-lnt+t^2/2+c and then turn back to x and get:

    -ln(cosx)+cos^2(x)/2+c .
     
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