## anonymous 5 years ago anti-derivative of SQRT(1+(9x- (x^3/3))^2)

1. amistre64

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

2. anonymous

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

3. amistre64

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

4. anonymous

yes

5. amistre64

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

6. anonymous

omg i hate calculus lol

7. amistre64

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

8. anonymous

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

9. amistre64

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

10. amistre64

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

11. amistre64

but it shows me another idea lol

12. anonymous

yea im not so good with substitution either

13. amistre64

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

14. amistre64

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

15. anonymous

ok im just trying to follow the process here

16. amistre64

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

17. anonymous

sure i couldnt go anymore wrong with this problem lol

18. amistre64

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

19. anonymous

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

20. amistre64

now we can factor under the radical to simply fy things

21. amistre64

sooo...how should we factor it :) all polynomials higher then degrees 2 can factor to linears and or irreducibla quads

22. anonymous

i honestly have no idea lol

23. amistre64

ima try some synthetic stuff and hope i get lucky

24. anonymous

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

25. amistre64

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

26. anonymous

whos that? some really smart kid im guessing :)

27. amistre64

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

28. anonymous

o gotcha lol..ok well maybe i went wrong somewhere..the question is find the length of the arch of the parabola y=9-x^2 that lies above the x-axis

29. amistre64

wolframalpha.com

30. amistre64

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

31. amistre64

you had the wrong stuff inputed lol

32. anonymous

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

33. amistre64

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

34. amistre64

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

35. anonymous

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

36. amistre64

dy/dx = -2x

37. anonymous

ooo i did the anti derivative didnt i ..i do that all the time because we keep having to switch from derivativeto anti derivative lol

38. amistre64

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

39. amistre64

integrals

40. anonymous

yea that what we do when we use integrals

41. amistre64

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

42. anonymous

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

43. anonymous

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

44. amistre64

easier than you would that other monstrocity lol

45. anonymous

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

46. amistre64

sqrt(1+tan^2(u)) sec^2(u) du ------------------------- 4 sec(u). sec^2(u) du ---------------- 4 sec^3(u) du ---------- 4

47. amistre64

got that outta whack a little

48. anonymous

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

49. amistre64