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Need help with 2 integrals.

  1. Mar 14, 2007 #1
    Hey, I have a crappy community college professor and apparently I am not smart enough to figure out the parts she didn't teach.

    the definite inegral of (ln x)^2 - 1 (part of a geometry problem)
    from e^-1 to e

    I tried this:

    [(x ln x - x)^3]/3 - x

    And got 2.48, but looking at the graph it looks like it should be about 1.5.

    This is just a homework problem, but none of the class can get it.

    I also have another question on a different problem.

    a definite integral from -.5 to .5

    SQRT[1-(2 cos (pi*x))^2]

    I got as far as SQRT[1 - 4pi^2sin^2(pi*x) and I have NO idea how to integrate that.

    None of the class could figure out this one either.

    My calculus teacher is about 75 years old, and she shouldn't be teaching, I feel like I haven't learned anything except what I have taught myself.

    Thank you for any help!
  2. jcsd
  3. Mar 14, 2007 #2
    Could you post the original problems?

    Also, I am not sure what you mean by:
    If you evaluate the integral you should get:

    [tex]\int_{e^{-1}}^{e}\left( (\ln x)^2 - 1 \right) dx \approx -1.47[/tex]

    Is that what you mean by 1.5? Do you need help evaluating the integral? What are you having trouble with?
  4. Mar 14, 2007 #3
    yeah, it looked by eye like it should about 1.5, but the equation that I got when I integrated, what I posted above, gave me about 2.5.

    I'm just not sure how to integrate the square of (log x)...
  5. Mar 14, 2007 #4
    If you can, use a table (there should be one in the back of your book). If you want to, or have to, do it by hand, try substitution and then use integration by parts (twice :smile:). There may be a more clever way to do it, but what I said works (and it is not that bad).
  6. Mar 15, 2007 #5

    Gib Z

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    [tex]\int\left( (\ln x)^2 - 1 \right) dx = x(\ln x -1)^2 + C[/tex]

    That makes your definite integral [tex]- 4(e^{-1})[/tex], which is approx
  7. Mar 15, 2007 #6
    thanks for the help!
  8. Mar 15, 2007 #7


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    Do you know how to part integrate ?If so, then your first integral shouldn't be difficult.

    [tex] \int_{1/e}^{e} \ln^{2} x \ {}dx -\int_{1/e}^{e}{}dx [/tex]

    =to be solved -(e-1/e).
  9. Mar 15, 2007 #8


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    Also, are you absolutely sure about the form of the second integral ? Cause i'm getting a nasty combination of elliptic integrals.
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