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Integration Issues a bit

  1. Jun 30, 2008 #1
    Integration Issues...a bit urgent

    Hey having problems with five of my assignment questions, any help would be appreciated.

    1. The problem statement, all variables and given/known data
    (problems attached)

    2. Relevant equations

    3. The attempt at a solution

    For the first one, I tried to solve it by using the sin-1 x as 'u' and integrating the u and getting u3 / 3 + c, but that doesn't seem to work. Then I thought I could use the Integration by parts formula, but since its (sin-1x)2, both u and v would be the same....

    Second one: I thought of taking the y0.5 as the substitution 'u', but it doesn't seem to make much difference.

    Third one: I thought I could seperate it into ln( (1+t)0.5) + ln (t0.5) and solve them individually...

    Fourth one: The improper fractions technique, but then I can't seem to get legit values for A, B and C

    Last one: Direct Comparison Test, comparing it to the fraction: x2 / (x+3)3

    Attached Files:

  2. jcsd
  3. Jun 30, 2008 #2
    Re: Integration Issues...a bit urgent

    Why would it matter if u and dv (not v) are the same? In fact, for 1, that would be probably the only way to do it.
  4. Jul 1, 2008 #3


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    Re: Integration Issues...a bit urgent

    I'll need to deal with this in parts. Here's what I suggest at the moment.

    1) It may be six-of-one, half-a-dozen-of-the-other as to whether you should use u = (arcsin x) or u = (arcsin x)^2 .

    If you do the former, you will have to integrate dv = (arcsin x) dx , which isn't too bad. It's tidier, though, to do the latter and have dv = dx.

    It looks like either way you will have to integrate

    [tex]\frac{x}{\sqrt{1 - x^2} } arcsin x dx [/tex]

    which means a second integration by parts with u = arcsin x and dv for the rest, which is a substitution integral. I think it does settle down after that... (Nasty little problem.)

    EDIT: Worked it out with u = (arcsin x)^2. The second integration-by-parts is easy.

    2) Try v = sqrt(y). You'll then have an integration-by-parts to do; the v du integral looks like it will require a trig substitution, but then you're done.

    4) Another multiple-technique problem. (I'm suspecting they all are: I think your instructor is out to get you...) Substitute u = e^x to create an integrand with a quadratic polynomial over the square root of a quadratic polynomial. Then try completing the square for the argument of the radical. It may be a trig-substitution integral from there...

    5) I take it you're supposed to see if this improper integral diverges, or else evaluate it if it converges. You're correct in that you want to focus on the [tex]\frac{x^2}{(x+3)^3}[/tex] term. The (sin x)^2 factor is there, I suspect, to attempt to foil you in this, since it is bounded above by 1, so it is "usually" smaller than [tex]\frac{x^2}{(x+3)^3}[/tex]. You may have to make some sort of argument that, averaged over a period, sine-squared is non-zero, so this integral term acts like a constant times 1/x , and thus diverges. (One certainly has an intuition that this thing diverges...)

    Still thinking about #3...
    Last edited: Jul 1, 2008
  5. Jul 1, 2008 #4
    Re: Integration Issues...a bit urgent

    for 3, your best bet would probably be to use integration by parts with u=1 and dv=... . Doing so will "get rid" of the ln under the integral sign.
  6. Jul 1, 2008 #5
    Re: Integration Issues...a bit urgent

    Ha, I remember doing the first four from my textbook.
    1) Use parts, let u=arcsin[x]^2 dv=dx. Should be clear what to do next.
    2) Use sub, let u=sqrt[x] so 2u du =dx, then use parts.
    Never could solve the others, didn't really care either because they never come up in the real world.
  7. Jul 1, 2008 #6


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    Re: Integration Issues...a bit urgent

    For 3, the solution I could come up with involves 2 substitutions and quite a number of integration by parts. As foxjwill said, do it first by integration by parts, then make a substitution for t^(1/2), you'll end up with something you should recognise easily as integrable by a trigo substitution, but it gets tedious from this point, though it's doable.

    EDIT: The online integrator's answer involves a hyperbolic sine, which strongly implies that you can shorten the number of steps needed by familiarising yourself with integrals and differentiation of hyperbolic trigo functions.
  8. Jul 3, 2008 #7

    Gib Z

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    Re: Integration Issues...a bit urgent

    That is correct, for the third one [itex] t = \sinh^2 x[/itex] is good to use, it comes out in a few lines if you are familiar with hyperbolic trig functions.
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