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how you wanna do this lol the easy way or the hard way

lets go with easy because this is the part of problems i can never do

\[\sqrt{1+(9x-\frac{x^3}{3})^2}\] lets start by making sure we got the right problem is this it?

yes

well; its gonna be a trig sub that turns to int(sec(x)) dx inthe end lol

omg i hate calculus lol

lol...trig sub just cleans it up thats all

yea but i integrated this on my calculator and i still got the wrong answer

\[\int_{} \sqrt{1+tan^2(u)}du\] ;\[u= \frac{27x -x^3}{3}\]

du = 27-3x^2/3 = 9-x^2 dx; maybe that aint a good idea lol

but it shows me another idea lol

yea im not so good with substitution either

u = 9x-x^3 ; du = 9-3x^2 dx
{S} sqrt(1+(u/3)^2) du
--------------
9-x^2
maybe?

let me stare blindly at it for a few moments and let my thoughts congeal lol

ok im just trying to follow the process here

we might be better of expanding and combining everything under the radical into one shot

sure i couldnt go anymore wrong with this problem lol

\[\frac{\sqrt{x^6 + 729x^2 -54x^4 +9}}{3}\] is what i get with that ...maybe

yea i got something completely off but u r def prolly right lol

now we can factor under the radical to simply fy things

i honestly have no idea lol

ima try some synthetic stuff and hope i get lucky

lol ok im attempting to do other problems..im going to fail this quiz tomorrow lol

well, the good news is; wolfram couldnt even figure it out lol

whos that? some really smart kid im guessing :)

www.wolframalpa.com type in:
int(sqrt(1+(9x-(x^3)/3)^2))dx

wolframalpha.com

.... why you wanna keep secrets from me; i cant read your mind

you had the wrong stuff inputed lol

i did?? well the equation we use is SQRT(1+(dy/dx)^2

\[\int\limits_{} \sqrt{1 + [f'x]^2} dx\]

and is the derivative of 9-x^2 = to the integral of 9-x^2?

yea and the derivative of 9-x^2 is 9x-(x^3/3) right?

dy/dx = -2x

anti derivative is a useless term; later chapters call them what they are inegrals lol

integrals

yea that what we do when we use integrals

{S} sqrt(1+4x^2) dx ; [a,b] whatever your interval s its up there some where

ok now i got it..wow i am retarded lol i do that a lot to

so now how to integrate SQRT(1+4x^2)

easier than you would that other monstrocity lol

i got 2/3(1+4x^2)^(3/2) and then do you also multiply that by the derivative of the inside??

got that outta whack a little

ok i g2g i got soccer..but thanks for the help :)

http://www.wolframalpha.com/input/?i=int%28+sqrt%28+1%2B+%28-2x%29^2%29%29+dx+from+-3+to+3