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When to use what the u and du substitutes

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I just want to know what to use for the u and du substitutes. I know the final and I can see a short cut to the answer, but I will need to show my work and trying the find the correct u and du is proving hard. The correct answer is 2/3sin^-1(3x)+C
  2. jcsd
  3. Jul 20, 2010 #2
    Re: U-substitution

    You said you saw a shortcut to the answer. What shortcut is that? There isn't much to this u-substitution, so I'm curious if you might just have the answer and not quite recognize it.
  4. Jul 20, 2010 #3
    Re: U-substitution

    It's not a short cut in the sense of ∫sinx=-cosx, just from working the problem, I see that in the equation I take the numerator,2, and pull it out, plus I know that 1/(1-x^2) = sin^-1(x). So I can take the √9 and get 3, so the answer is 2/3sin^-1(3x)+C. But like I said, I need to show the work, and when I take (1-9x^2) for u I have no other x to account for in du. So what would my u and du be?
  5. Jul 20, 2010 #4
    Re: U-substitution

    I don't think you fully understand what you're doing to get from the original question to the solution.

    I'll take a shot at what I think u and du are, but can you please finish the question showing your work from start to finish for us? I want to make sure you fully understand what you're doing.

    Why not try,

    [tex] u = 3x \ldots \frac{du}{dx} = 3 [/tex]

  6. Jul 20, 2010 #5


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    Re: U-substitution

    How about sin(u)^2=9*x^2? You have to figure out the du.
  7. Jul 20, 2010 #6
    Re: U-substitution

    That did it. Thanks


    u^2=9x^2 u=3x & du=3*dx 1/3du*dx




    and I had another one that I could see the answer, but couldn't find u and du till this help and can solve on paper.
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