So I have a square root integral as follows...

x*sqrt((x^2) + 3)dx

Do I take the u-sub of this thing?

like...

u = x^2

du = 2xdx

xdx = du/2

and what do I do from there?

I know that it will be sqrt(u + 3)du/2

but that doesnt help me much...

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- Apr 25th 2010, 12:32 AMRET80basic integral refresher with a square root
So I have a square root integral as follows...

x*sqrt((x^2) + 3)dx

Do I take the u-sub of this thing?

like...

u = x^2

du = 2xdx

xdx = du/2

and what do I do from there?

I know that it will be sqrt(u + 3)du/2

but that doesnt help me much... - Apr 25th 2010, 12:34 AMMacstersUndead
Try substituting x^2 + 3 instead

u = x^2 + 3

du = 2x dx

du/2 = x dx

so the resulting integral will be sqrt(u)/2 - Apr 25th 2010, 12:38 AMmathemagister
- Apr 25th 2010, 12:42 AMRET80
Well I should say

The equation actually is...

in terms of rdrd(theta)

and I seperated theta and r out into seperate integrals

I have r*sqrt((r^2) - (a^2))dr

the equation was originally derived from the equation of a sphere...

z^2 = x^2 + y^2 - a^2

and a is some arbitrary radius. - Apr 25th 2010, 02:52 AMHallsofIvy
Yes, and both MacstersUndead and Mathmagister have told you what to do: let $\displaystyle u= r^2- a^2$.