anonymous 5 years ago Integral of 1/(sqrt(9-4x^2))

1. Owlfred

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!

2. anonymous

use wolfram alpha

3. anonymous

It doesn't show any of the steps.

4. anonymous

it does, click show steps

5. nowhereman

Substitute $$\frac{2}{3}x = \sin z$$

6. anonymous

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

7. 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

8. 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.

9. anonymous

yes it does show the steps

10. anonymous

its one step

11. 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.

12. anonymous

its one ugly giant step

13. anonymous

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

14. anonymous

check pauls notes on trig substitution, whatever

15. nowhereman

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

16. nowhereman

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

17. anonymous

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

18. anonymous

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

19. nowhereman

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

20. 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.

21. anonymous

multiply*

22. 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.

23. 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.

24. anonymous

ok

25. 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 $$[0, 2]$$

26. anonymous

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

27. anonymous

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