Question

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

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

use wolfram alpha

Answer 3

It doesn't show any of the steps.

Answer 4

it does, click show steps

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

yes it does show the steps

Answer 10

its one step

Answer 11

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.

Answer 12

its one ugly giant step

Answer 13

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

Answer 14

check pauls notes on trig substitution, whatever

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

multiply*

Answer 22

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.

Answer 23

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.

Answer 24

ok

Answer 25

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 \([0, 2]\)

Answer 26

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

Answer 27

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