# basic integral refresher with a square root

• Apr 25th 2010, 12:32 AM
RET80
basic 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 AM
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 AM
mathemagister
Quote:

Originally Posted by RET80
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...

Well, you could stll do it from there, but try using $u=x^2+3$.

$du = 2x \; dx$

$\frac{1}{2}\; du = x \; dx$

You should find the rest follows through.
Hope that helps :)

Mathemagister
• Apr 25th 2010, 12:42 AM
RET80
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 AM
HallsofIvy
Yes, and both MacstersUndead and Mathmagister have told you what to do: let $u= r^2- a^2$.