Integral of 1/(sqrt(9-4x^2))

- anonymous

Integral of 1/(sqrt(9-4x^2))

- katieb

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- Owlfred

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- anonymous

use wolfram alpha

- anonymous

It doesn't show any of the steps.

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- anonymous

it does, click show steps

- nowhereman

Substitute \(\frac{2}{3}x = \sin z\)

- anonymous

nowhereman, I got that part. My only problem is getting the 1/2 in front of the integral.

- 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

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

yes it does show the steps

- anonymous

its one step

- 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

its one ugly giant step

- anonymous

then pay for a tutor, , and dont tell me what to do

- anonymous

check pauls notes on trig substitution, whatever

- nowhereman

laplace, could you please post your question seperately, I will help knightfire

- nowhereman

Could you show the substitution step you are doing knightfire, that's where the 1/2 comes from.

- anonymous

So I set x = 3/2(sinz), dx = 3coszdz, and sqrt(9-4x^2) = 3cosz

- anonymous

I plug them in for their appropriate parts and I get z +c

- nowhereman

aha, that's the mistake, dx = 3/2 cos z dz

- 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

multiply*

- 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

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

ok

- 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

\[\int\limits{1/\sqrt{(9-4x ^{2})}}dx\]
\[=1/2 \int\limits 1/\sqrt{((3/2)^{2}-x ^{2})} dx\]
\[= (1/2)*2/(2*3) \ln ((3/2+x)/(3/2-x)) +c\]
\[=1/6 \ln ((3+2x)/(3-2x))+c\]

- anonymous

\[\int\limits \frac{1}{\sqrt{9 - 4*x^2}} \, dx=\frac{1}{2} \text{ArcSin}\left[\frac{2 x}{3}\right]+c \]

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