# Integration w/ u-substitution, work is inside

• Oct 9th 2009, 07:14 AM
janedoe
Integration w/ u-substitution, work is inside
http://i36.tinypic.com/rsas13.jpg

Thanks in advance for any help!
• Oct 9th 2009, 07:22 AM
pflo
Why use u-substitution on this problem?
$\displaystyle \int{t^2(t-\frac{2}{t}) dt}$
$\displaystyle \int{(t^3-2t)dt}$
$\displaystyle \frac{t^4}{4}-t^2+c$

I suppose if you HAVE to use u-substitution, you could let $\displaystyle u=t^2$ instead.
That would give you the following integral:
$\displaystyle \int{(\frac{u}{2}-1)du}$
• Oct 9th 2009, 07:24 AM
janedoe
Quote:

Originally Posted by pflo
Why use u-substitution on this problem?

Well, that's what the lesson was on, so I just assumed we had to do it that way. But yeah, that's easier. Does that mean u-sub would be nearly impossible/very difficult if I were to do it?
• Oct 9th 2009, 07:32 AM
pflo
Quote:

Originally Posted by janedoe
Well, that's what the lesson was on, so I just assumed we had to do it that way. But yeah, that's easier. Does that mean u-sub would be nearly impossible/very difficult if I were to do it?

I edited my original reply to include a method where you could use u-substitution.

One note: When using u-substitution, always look for the derivative of u (the one with the original variable) to cancel out. In this case, you identified the problem with the substitution you originally did - couldn't cancel out the t's. But with $\displaystyle u=t^2$ the derivative would include a 't' which WOULD cancel out.