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Math Help - difficulty with integral: x over square root

  1. #1
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    difficulty with integral: x over square root

    Hi. Could someone please help me figure out this integral, which I am doing for extra practice. (I'm doing terribly on integrals overall, and hoping some extra practice will help.) I need to get the integral of this function:

    x / sqrt (9 - x^2)

    Because of the chapter this problem is in, I think it is solved working with u-substition and the integral of the natural logarithm function, but I can't seem to make it work.
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    Re: difficulty with integral: x over square root

    Quote Originally Posted by infraRed View Post
    Hi. Could someone please help me figure out this integral, which I am doing for extra practice. (I'm doing terribly on integrals overall, and hoping some extra practice will help.) I need to get the integral of this function:

    x / sqrt (9 - x^2)

    Because of the chapter this problem is in, I think it is solved working with u-substition and the integral of the natural logarithm function, but I can't seem to make it work.
    You have a square root in the denominator. That usually means a trig substitution at some point.

    I'd start with the substitution u = 3x. Then simplify the integrand. Then use the trig substitution u = sin(y) and see where things lead you.

    -Dan
    Last edited by topsquark; January 26th 2013 at 04:15 PM.
    Thanks from infraRed
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    Re: difficulty with integral: x over square root

    Thank you for your assistance. I must admit, however, that I'm not quite following you. If u = 3x, what does it replace?
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    Re: difficulty with integral: x over square root

    Quote Originally Posted by infraRed View Post
    the integral of this function:
    x / sqrt (9 - x^2)

    Let u=9-x^2 then du=-2xdx so

    \int {\frac{{xdx}}{{\sqrt {9 - x^2 } }} = \frac{{ - 1}}{2}\int {\frac{{du}}{{\sqrt u }}} }
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    Re: difficulty with integral: x over square root

    Quote Originally Posted by Plato View Post
    Let u=9-x^2 then du=-2xdx so

    \int {\frac{{xdx}}{{\sqrt {9 - x^2 } }} = \frac{{ - 1}}{2}\int {\frac{{du}}{{\sqrt u }}} }
    Ah, and then (-1/2) * 2u^(1/2) = -u^(1/2) = -sqrt(9-x^2). It seems so simple now. I was expecting the logarithm aspect to be in the somehow... guess I over-fixated on it.

    By the way, does MHF have a quick tutorial page / cheatsheet for that TEX graphics stuff you guys are all doing?
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    Forum Admin topsquark's Avatar
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    Re: difficulty with integral: x over square root

    Quote Originally Posted by Plato View Post
    Let u=9-x^2 then du=-2xdx so

    \int {\frac{{xdx}}{{\sqrt {9 - x^2 } }} = \frac{{ - 1}}{2}\int {\frac{{du}}{{\sqrt u }}} }
    Ah, you found a less complicated way. Thanks!

    Quote Originally Posted by infraRed View Post
    Ah, and then (-1/2) * 2u^(1/2) = -u^(1/2) = -sqrt(9-x^2). It seems so simple now. I was expecting the logarithm aspect to be in the somehow... guess I over-fixated on it.

    By the way, does MHF have a quick tutorial page / cheatsheet for that TEX graphics stuff you guys are all doing?
    Take a look at our "Latex" forum. It doesn't take to long to learn how to do the simple stuff.

    -Dan
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