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anonymous

  • one year ago

Integrate ((9sin^4(t) + 9cos^4(t))^1/2

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  1. anonymous
    • one year ago
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    It's all underneath the square root

  2. Luigi0210
    • one year ago
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    This right? \(\Large \int (\sqrt{9sin^4 t +9cos^4 t}) dt\)

  3. anonymous
    • one year ago
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    Yes

  4. Luigi0210
    • one year ago
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    I would trying reducing it down, factor out that 9 first: \(\Large \int \sqrt{9(sin^4 t +cos^4 t}) dt \) Since 9 can come out of the square roots, you get this: \(\Large \int 3(\sqrt{sin^4 t +cos^4 t}) dt \) Make sense so far?

  5. anonymous
    • one year ago
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    Yep got that down already - then I tried to use half angle identities but got stuck there

  6. Luigi0210
    • one year ago
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    You factored a square from each right?

  7. Luigi0210
    • one year ago
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    \(\Large 3\int (\sqrt{(sin^2 t )^2+(cos^2 t)^2} dt \) ?

  8. anonymous
    • one year ago
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    Yep then used 1+cos(2t)/2 and 1-cos(2t)/2

  9. anonymous
    • one year ago
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    ???

  10. Luigi0210
    • one year ago
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    But yea, what did you get after you tried that?

  11. anonymous
    • one year ago
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    Tried expanding them - but since its a square under square root can be factored out?

  12. anonymous
    • one year ago
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    ??

  13. EmmaTassone
    • one year ago
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    i dont want to be killjoy but i think that integral dont have solution. its similar to:\[\int\limits_{}^{}\sqrt{a.\cos^{2}(x)+b.\sin ^{2}(x)} And this integral dont have solution neither.Its called the elliptic integral.

  14. Luigi0210
    • one year ago
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    Yea that's what I was getting too

  15. Luigi0210
    • one year ago
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    And your num lock is off or something there? :P

  16. anonymous
    • one year ago
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    I am doing arc length it's from 0 to pi/2

  17. Luigi0210
    • one year ago
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    That would of been a bit helpful to know

  18. anonymous
    • one year ago
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    Answer in back says 3/2

  19. EmmaTassone
    • one year ago
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    Maybe its a numerical solution or another way to calculate it, because you cant calculate it direclty with integrating methods.

  20. IrishBoy123
    • one year ago
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    you should post or scan the question it looks like an arc lngth, but why guess? :p

  21. amistre64
    • one year ago
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    might help to post the actual question ... or scan yeah

  22. anonymous
    • one year ago
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    Ok will try to write it out am on my iPad so here it goes Find the length of the following r(t) = <cos^3(t), sin^3(t)

  23. anonymous
    • one year ago
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    So found derivative and then did magnitude and got to where the question is now !! :)

  24. anonymous
    • one year ago
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    :-/

  25. amistre64
    • one year ago
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    what techniques have you covered?

  26. anonymous
    • one year ago
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    I had thought to use the half angle formula to then take integral of ! In class covered take r'(t) then magnitude of r'(t) then take integral

  27. EmmaTassone
    • one year ago
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    i think you have derivated wrong, check it again

  28. anonymous
    • one year ago
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    Cos^3(t) = -3sin^2(t) Sin^3(t) = 3cos^2(t)

  29. amistre64
    • one year ago
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    im thinking .. and it may not be any better y = sin^3(acos(cbrt(x))) to eliminate the parameter maybe?

  30. EmmaTassone
    • one year ago
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    you are forgeting chain rule

  31. amistre64
    • one year ago
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    accursed chain rule lol

  32. amistre64
    • one year ago
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    writing it as say: (cos(t))^3 might make it easier to recall chaining it

  33. anonymous
    • one year ago
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    Then what is it !? There is nothing with the t to do chain rule

  34. amistre64
    • one year ago
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    if u = cos(t) u^3 derives to 3u^2 u'

  35. EmmaTassone
    • one year ago
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    \[(\cos ^{3}(x))\prime=3.\cos ^{2}(x).(-\sin(x))\]

  36. EmmaTassone
    • one year ago
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    same with the sine

  37. anonymous
    • one year ago
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    So sine = 3sin^2(t)cos(t)

  38. anonymous
    • one year ago
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    Gosh dang chain rule lol

  39. EmmaTassone
    • one year ago
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    yep, it was a calculation mistake lol

  40. anonymous
    • one year ago
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    Wow lol such stupid mistake lol

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