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Help! i beg of you

  1. Jun 26, 2004 #1
    i for the life of me can't figure out how to integrate this:

    [tex] \int \sin{(x^{1/2})} dx [/tex]
  2. jcsd
  3. Jun 26, 2004 #2
    Make the substitution [tex]u = \sqrt{x}[/tex] which will turn the integral into something like [tex]\int \left(\sin{u} \times 2\sqrt{x}\right) du[/tex]. Apply the substitution again and use integration by parts...
    Last edited: Jun 26, 2004
  4. Jun 26, 2004 #3
    We don't have enough x's in there, so use integration by parts to get some. Let u = sin(x^(1/2)) and dv = dx. After that, you'll get an integral that can be solved through a substitution and then a couple of applications of by parts.

  5. Jun 26, 2004 #4
    thanks, got it now
  6. Jun 29, 2004 #5
    Ok a new integral I can’t figure out how to take

    [tex] \int \frac{1}{4y^{2} - 4y - 3} dy[/tex]

    Any suggestions?
  7. Jun 29, 2004 #6
    Partial fractions.

  8. Jun 30, 2004 #7
    Oops I posted the wrong integral, I meant:
    [tex] \int \frac{1}{(4y^{2} - 4y - 3)^{1/2}} dy[/tex]
  9. Jun 30, 2004 #8


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    Try completing the square and using a substitution of what you get inside the ()^2 bit. It should then be a simple matter of knocking it into standard form.
  10. Jun 30, 2004 #9
    Try making a change till you get the folowing formula
    [tex] \int \frac{1}{((Y-A)^2+B^2)^{1/2}} dY[/tex]
    Sorry I forgot the formula of how to go on, you can check it out in your textbooks.
  11. Jun 30, 2004 #10


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    This might help (a standard form)

    [tex]\int \frac{1}{(x^2 + a^2)^{\frac{1}{2}}} dx = \ln \left(x + \sqrt{x^2 + a^2} \right)[/tex]

    Though it would appear to me you are on the wrong track. Have checked your integral out at http://integrals.wolfram.com (you need to check their syntax on how to input function it's a little temperamental)
  12. Jun 30, 2004 #11
    My suggestion is complete the square on the denominator. Than it should be a matter of using a trig substitution. Looks to me like it will be a secant substitution.
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