difficulty with integral: x over square root

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January 26th 2013, 02:50 PM
infraRed

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.
January 26th 2013, 03:40 PM
topsquark

Re: difficulty with integral: x over square root

Quote:

Originally Posted by infraRed

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
January 26th 2013, 04:11 PM
infraRed

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?
January 26th 2013, 05:28 PM
Plato

Re: difficulty with integral: x over square root

Quote:

Originally Posted by infraRed

the integral of this function:
x / sqrt (9 - x^2)

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

$\int {\frac{{xdx}}{{\sqrt {9 - x^2 } }} = \frac{{ - 1}}{2}\int {\frac{{du}}{{\sqrt u }}} }$
January 26th 2013, 05:50 PM
infraRed

Re: difficulty with integral: x over square root

Quote:

Originally Posted by Plato

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?
January 27th 2013, 04:28 PM
topsquark

Re: difficulty with integral: x over square root

Quote:

Originally Posted by Plato

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

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