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Using u substitution for integrating.

  1. May 22, 2014 #1
    So I am pretty bad at u substitution.
    I don't really get how to replace values with du or u.

    Can you please give me tips on how to do u substitution well?
    Thanks.
     
  2. jcsd
  3. May 23, 2014 #2
    In the most basic example, u corresponds to some expression while du corresponds to the derivative of that expression. The main idea is to look for this pattern and eventually integrate.

    Example:

    [itex]1. \int 2x \sqrt{x^2+4} \: dx[/itex]

    Let
    [itex]u = x^2+4[/itex], then [itex]du = 2x \: dx[/itex], thus you have the form

    [itex]\int du \sqrt{u} \: [/itex] or simply [itex]\int du \: u^{1/2} \: [/itex], and then you integrate:

    [itex]\int 2x \sqrt{x^2+4} \: dx = 2/3(x^2+4)^{3/2} + C[/itex]

    The best tip I could give you is to practice a lot and try to identify these patterns quickly. However, it may not be as easy and obvious at first glance, and sometimes algebraic manipulation or other things may come in handy before integrating. If it would be a [itex]4x[/itex] instead of a [itex]2x[/itex] then what would you do?

    More examples:

    [itex]2. \int \frac{ln(x)}{x}\: dx[/itex]

    [itex]3. \int {sin(x)}^{3} \: cos(x) \: dx[/itex]
     
    Last edited: May 23, 2014
  4. May 24, 2014 #3
    So... for number 2:
    du = 1/x dx
    and u = ln(x)
    ∫u du =
    (u^2)/2 + c
    ([ln(x)]^2)/2 + c
     
  5. May 24, 2014 #4
    But I don't know how to evaluate number 3.
    what about sin^3(x)?
     
  6. May 24, 2014 #5

    micromass

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    Try a substitution ##u=\cos(x)##. That will work rather nicely if you know your trig.
     
  7. May 28, 2014 #6
    Often times I find it's useful to just try things, as there are only so many choices you can make. Sometimes they tend to work themselves out if you're lucky. Sure, this might seem fairly reckless, but it helps build your intuition when you see how it's wrong.
     
  8. May 28, 2014 #7
    Or how about ##u=\sin x##? :rolleyes:
     
  9. May 29, 2014 #8

    HallsofIvy

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    Either will work- but, for integrating [itex]sin^2(x)cos(x)[itex], u= sin(x) is much easier than u= cos(x)!
     
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