Stuck On An Integral: (9-3x)^3 - Help!

[ns5032]

In the middle of doing an integral for Calc 3, and I came across (9-3x)^3, and completely forgot how to do this from algebra! Help!

 

[rock.freak667]

Use u=9-3x
du=-3 dx so that -du/3 = dx

so you must really integrate

$$\int \frac{-1}{3}u^3 du$$

 

[ns5032]

Oh, I guess I didn't phrase my question right, huh? Haha. I'm doing an integral, yes, but I'm not actually taking the integral of that... I just need to simply that part of a loooong answer I got inside an integral... if that makes sense. Just algebra stuff.

 

[ns5032]

Is this what it is:

(a + b)^3 = a3 + 3a^2b + 3ab^2 + b^3

?

 

[rock.freak667]

yes that is the expansion of (a+b)^3