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Another Part where my brain's Fried |
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| May7-05, 02:52 PM | #1 |
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Another Part where my brain's Fried
[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. |
| May7-05, 03:12 PM | #2 |
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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... |
| May7-05, 03:13 PM | #3 |
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Damn it, no wonder I couldn't get it to work (I tried x = 2cos u)!
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| May7-05, 03:13 PM | #4 |
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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. |
| May7-05, 03:23 PM | #5 |
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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 "-". |
| May7-05, 03:26 PM | #6 |
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| May8-05, 11:18 AM | #7 |
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Thanks once more! :)
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