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  • one year ago

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  1. anonymous
    • one year ago
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    How may I solve the following integral? \[\int\limits (16-x^2)^{\frac{3}{2}}dx\]

  2. anonymous
    • one year ago
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    I know for \[\int\limits (a^2-x^2)^\frac{1}{2}dx=\frac{x}{2}.(a^2-x^2)^\frac{1}{2}+\frac{a^2}{2}\sin^{-1}(\frac{x}{a})\] But this is cubed now

  3. Loser66
    • one year ago
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    use trig substitute

  4. Loser66
    • one year ago
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    x = 4 sint dx = 4 cost dt x^2 = 16 sin^2 t 16 -x^2 = 16 cos^2 t

  5. Loser66
    • one year ago
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    (16 -x^2)^3/2 = 4^3 cos^3t ok?

  6. Loser66
    • one year ago
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    no?? hehehe...

  7. anonymous
    • one year ago
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    \[I=256 \int\limits \cos^4(t)dt\]

  8. Loser66
    • one year ago
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    ok, go ahead

  9. anonymous
    • one year ago
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    thanks a lot, let me try it now

  10. anonymous
    • one year ago
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    Ok so I'm getting a really weird answer \[I=96 \sin^{-1}({\frac{x}{4}})+8x\sqrt{16-x^2}+\frac{x\sqrt{16-x^2}(8-x^2)}{32}\] Now I can't say for sure how much accurate this answer is, but more or less I now understand how to do that integral thanks @Loser66

  11. anonymous
    • one year ago
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    and +C of course

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