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anonymous

  • 5 years ago

whats the definite integral from pi/6 to pi/4 of -x/(1-x^4)^1/2

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  1. anonymous
    • 5 years ago
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    You cannot do this by hand because letting u=1-x^4 its derivative would equate -4x^3 dx, which cannot be substituted into the equation. To find the result, type in the equation into y= on your calculator. Then go 2ND->Calc (top row of buttons) and hit #7 for integration. It will graph the function, and then ask for a lower and upper bound which is obviously the limits you provided.

  2. anonymous
    • 5 years ago
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    ok, thanks a lot

  3. myininaya
    • 5 years ago
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    did you try a trig substitution? that might work

  4. anonymous
    • 5 years ago
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    that's not true, it's a identity it's the inverse sin identity.

  5. anonymous
    • 5 years ago
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    thats what i'm supposed to do, but i'm extremely lost. What do you mean it's the inverse sin identity?

  6. anonymous
    • 5 years ago
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    \[-\frac{1}{2}\sin^{-1} (x^2)dx\] from pi/6 to pi/4

  7. anonymous
    • 5 years ago
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    you can derive it, but most teachers (even in college) do not require it most teachers don't even require you to know them but some want you to memorize them that's the only way to recognize them that integral is the derivative of the inverse sin function

  8. anonymous
    • 5 years ago
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    Oh god lol I feel like a dolt. Yes it's an inverse trig ID. Sorry

  9. myininaya
    • 5 years ago
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  10. anonymous
    • 5 years ago
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    whoa myininaya, did you just do that? lols props!

  11. anonymous
    • 5 years ago
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    i dont think we're there yet. we're supposed to substitute and let u=x^2

  12. myininaya
    • 5 years ago
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    yes lol

  13. anonymous
    • 5 years ago
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    then how are you going to integrate the u^2 without a u on the outside...

  14. myininaya
    • 5 years ago
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    they are not hard to derive. it is harder for me to memorize more stuff so i derive it everytime

  15. myininaya
    • 5 years ago
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    ? what u?

  16. anonymous
    • 5 years ago
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    i dont know, thats why im so confused

  17. anonymous
    • 5 years ago
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    myininaya- the hint that came with the question said to let u=x^2

  18. anonymous
    • 5 years ago
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    it's going to get really messy...but you can try u - subbing again....

  19. myininaya
    • 5 years ago
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    why let u=x^2 when you can let cos( pheta)=x^2

  20. myininaya
    • 5 years ago
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    where is the hint you guys are talking about? don't do that way. lol. hints can be gross

  21. anonymous
    • 5 years ago
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    LOLOLOLOLL go myininaya

  22. myininaya
    • 5 years ago
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    i mean we can try, but I don't see why

  23. anonymous
    • 5 years ago
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    its number 2 on the first page

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  24. myininaya
    • 5 years ago
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    have you done trig substitution before?

  25. anonymous
    • 5 years ago
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    nope

  26. myininaya
    • 5 years ago
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    ok let me think about then

  27. anonymous
    • 5 years ago
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    ok, thanks for helping me with this

  28. anonymous
    • 5 years ago
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    i think i understand what your book wants it want's u sub and trig id give me a min to type this out

  29. anonymous
    • 5 years ago
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    \[-\int\limits_{\pi/6}^{\pi/4}\frac{x}{\sqrt{1-x^4}}dx\] substitute u = x^2 and du = 2xdx \[-\frac{1}{2}\int\limits_{}^{}\frac{1}{\sqrt{1-u^2}}du\] now you use the trig id \[\int\limits_{}^{}\frac{1}{\sqrt{1-u^2}} = \sin^{-1} (u)\] so the answer is \[-\frac{1}{2}\sin^{-1} (u)\] sub x^2 back in \[-\frac{1}{2}\sin^{-1} (x^2)\]

  30. myininaya
    • 5 years ago
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    i don't see how we can do this without trig identity or trig substitution. You have to know something about those trig integral stuff to do this problem

  31. myininaya
    • 5 years ago
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    right

  32. anonymous
    • 5 years ago
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    doing it this way satisfies the hint of using u-sub ^_^

  33. anonymous
    • 5 years ago
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    ok, so then what do i do with the limits? f(b)-f(a)?

  34. myininaya
    • 5 years ago
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    just plug them into the antiderivative

  35. anonymous
    • 5 years ago
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    yup, since, we plugged x^2 back into u you just have to plug pi/6 and pi/4 into the answer that we got up there i stopped putting them in i'm sorry, i got lazy

  36. myininaya
    • 5 years ago
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    putting them is the easy part

  37. anonymous
    • 5 years ago
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    haha dont apologize, thanks for all your help, you guys saved me :)

  38. anonymous
    • 5 years ago
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    cheers good luck

  39. myininaya
    • 5 years ago
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    deriving them is fun too so if you don't want to memorize junk learn to derive them

  40. anonymous
    • 5 years ago
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    lols the smart one is to learn to derive them, that way, you can never really forget because if you forget, just derive it again :D

  41. anonymous
    • 5 years ago
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    hahaha i'll remember that

  42. myininaya
    • 5 years ago
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    lol right. i have to derive alot of stuff because my memory sucks

  43. anonymous
    • 5 years ago
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    are you an engineering major?

  44. myininaya
    • 5 years ago
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    no im a math major

  45. anonymous
    • 5 years ago
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    fun stuff, good luck with everything

  46. anonymous
    • 5 years ago
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    i'm computer engineering major

  47. anonymous
    • 5 years ago
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    got one right :D

  48. myininaya
    • 5 years ago
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    computer engineering is dreamy im going to start my phd next semester and i picked to do a research in the theory of cryptography

  49. anonymous
    • 5 years ago
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    that sounds interesting and hard at the same time, good luck

  50. anonymous
    • 5 years ago
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    dreamy? what does that mean? and i like cryptography, at least i think i do...it's problem solving yes? i'm a freshman btw lols

  51. myininaya
    • 5 years ago
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    lol i think i want to date it lol

  52. anonymous
    • 5 years ago
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    lololls i love programming which involves a lot of problem solving i've only taken one programming class in college so far, and it wasn't special... i mean compared to the rest of my classes, it was awesome, but .. ehh -_-

  53. anonymous
    • 5 years ago
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    which class? I'm a Computer Science major, love programming.

  54. anonymous
    • 5 years ago
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    i learned matlab it's basically just a super calculator there are a lot of stuff that you can do with it but it's not really a language...

  55. anonymous
    • 5 years ago
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    oh matlab is nuts. Used a ton for scientific programming.

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