Another Part where my brain's Fried


by Chaz706
Tags: brain, fried
Chaz706
Chaz706 is offline
#1
May7-05, 02:52 PM
P: 13
[tex] \int \sqrt{4-x^2}^3 dx [/tex]

I'm thinking integrating by parts would work, with u being that root and dv being dx, but is that the right method and direction. I've tried it and it seems more complicated.
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z0r
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#2
May7-05, 03:12 PM
P: 12
Use [tex]x = 2 \sin \theta [/tex]...
then,
[tex]dx = 2 \cos \theta d\theta [/tex],
and your integral becomes:
[tex]16 \int \cos^4 \theta d\theta [/tex] .
Use [tex]\cos{2\theta} = 2\cos^2 \theta - 1 [/tex], etc...
Nylex
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#3
May7-05, 03:13 PM
P: 554
Damn it, no wonder I couldn't get it to work (I tried x = 2cos u)!

dextercioby
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#4
May7-05, 03:13 PM
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Another Part where my brain's Fried


So it's

[tex]\int \left(4-x^{2}\right)^{\frac{3}{2}} \ dx[/tex]

How about a substitution

[tex] x=2\sin t [/tex] ?

Daniel.

EDIT:Didn't see the other posts.
HallsofIvy
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#5
May7-05, 03:23 PM
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x= 2cos(u) should work exactly like "x= 2 sin(u)": dx= -2 sin u du and
[tex]\sqrt{4- x^2}= \sqrt{4-4 cos^2(u)}= 2 sin(u)[/tex] so the integral becomes
[tex]-16\int sin^4(u) du[/tex] and the only difference is that "-".
Nylex
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#6
May7-05, 03:26 PM
P: 554
Quote Quote by HallsofIvy
x= 2cos(u) should work exactly like "x= 2 sin(u)": dx= -2 sin u du and
[tex]\sqrt{4- x^2}= \sqrt{4-4 cos^2(u)}= 2 sin(u)[/tex] so the integral becomes
[tex]-16\int sin^4(u) du[/tex] and the only difference is that "-".
Oops, I made a slight mistake!
Chaz706
Chaz706 is offline
#7
May8-05, 11:18 AM
P: 13
Thanks once more! :)


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