anonymous
  • anonymous
Integral of 1/(sqrt(9-4x^2))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Owlfred
  • Owlfred
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anonymous
  • anonymous
use wolfram alpha
anonymous
  • anonymous
It doesn't show any of the steps.

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anonymous
  • anonymous
it does, click show steps
nowhereman
  • nowhereman
Substitute \(\frac{2}{3}x = \sin z\)
anonymous
  • anonymous
nowhereman, I got that part. My only problem is getting the 1/2 in front of the integral.
anonymous
  • anonymous
nowhereman, revolve the region bounded by y = (x+2)^2 , y = 2, y=0, about the y axis (using shell method). i get a negative v0lume
anonymous
  • anonymous
laplace...http://i301.photobucket.com/albums/nn63/knightfire8/Screenshot2011-05-26at104452PM.png It doesn't show the steps first of all. Second..Im trying to ask a simple question, please dont use my question as your own space.
anonymous
  • anonymous
yes it does show the steps
anonymous
  • anonymous
its one step
anonymous
  • anonymous
It's definitely not one step. I have at least 4 to get to my answer which has everything except the 1/2. Now please move on.
anonymous
  • anonymous
its one ugly giant step
anonymous
  • anonymous
then pay for a tutor, , and dont tell me what to do
anonymous
  • anonymous
check pauls notes on trig substitution, whatever
nowhereman
  • nowhereman
laplace, could you please post your question seperately, I will help knightfire
nowhereman
  • nowhereman
Could you show the substitution step you are doing knightfire, that's where the 1/2 comes from.
anonymous
  • anonymous
So I set x = 3/2(sinz), dx = 3coszdz, and sqrt(9-4x^2) = 3cosz
anonymous
  • anonymous
I plug them in for their appropriate parts and I get z +c
nowhereman
  • nowhereman
aha, that's the mistake, dx = 3/2 cos z dz
anonymous
  • anonymous
Wait, so how did you get the 1/2 you multiple by? That's the part Im confused on. Because as mentioned before I got everything the same as the answer it listed except for 1/2.
anonymous
  • anonymous
multiply*
nowhereman
  • nowhereman
Because if you differentiate \(x\) you will get \(\mathrm{d}(\frac{3}{2}\sin z)/\mathrm{dz} = \frac{3}{2} \cos z\) and that's where your \(1/2\) is missing.
anonymous
  • anonymous
Ohhh ok that makes sense now. I guess I forgot that x and dx were directly related to each other with all these trig equations going on...Thanks so much nowhereman, you were most helpful.
anonymous
  • anonymous
ok
nowhereman
  • nowhereman
laplace, you might get a negative volume, because the area you are rotation is on the left side of the y-axis. Try to mirror it, i.e. you \(y \leq (x-2)^2\) and the interval \([0, 2]\)
anonymous
  • anonymous
\[\int\limits{1/\sqrt{(9-4x ^{2})}}dx\] \[=1/2 \int\limits 1/\sqrt{((3/2)^{2}-x ^{2})} dx\] \[= (1/2)*2/(2*3) \ln ((3/2+x)/(3/2-x)) +c\] \[=1/6 \ln ((3+2x)/(3-2x))+c\]
anonymous
  • anonymous
\[\int\limits \frac{1}{\sqrt{9 - 4*x^2}} \, dx=\frac{1}{2} \text{ArcSin}\left[\frac{2 x}{3}\right]+c \]

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