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anonymous

  • 5 years ago

integrate sin^43x dx

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  1. anonymous
    • 5 years ago
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    \[\int\limits \sin^{43}(x) dx= - \cos^{43}(x)\] honestly, this is a weird question lol, so I'm not sure? ._.

  2. anonymous
    • 5 years ago
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    + C ofcourse

  3. anonymous
    • 5 years ago
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    sstar, i don't know the answer but i'm almost sure that's not correct ><

  4. anonymous
    • 5 years ago
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    LOL >_< I know, it looks weird

  5. anonymous
    • 5 years ago
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    hmm, is it possible to let u = 43? I've never solved such thing lol

  6. anonymous
    • 5 years ago
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    meh... no...this method is funny lol

  7. anonymous
    • 5 years ago
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    sstarica the question is\[\int\limits \sin ^{4}3x dx\]

  8. anonymous
    • 5 years ago
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    no, you have to substitute (cos x)^2 = 1 - (sin x)^2

  9. anonymous
    • 5 years ago
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    oh that makes everything a lot better.... oneprince USE PARENTHESIS!

  10. anonymous
    • 5 years ago
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    ok chia

  11. anonymous
    • 5 years ago
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    no wait a min, isn't the sin to the power of 43?

  12. anonymous
    • 5 years ago
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    chia.. solve it

  13. anonymous
    • 5 years ago
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    was it to the power of 43? or to the power of 2? lol

  14. anonymous
    • 5 years ago
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    give me a sec

  15. anonymous
    • 5 years ago
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    it's sin(3x)^4

  16. anonymous
    • 5 years ago
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    that's the question?

  17. anonymous
    • 5 years ago
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    oh lol

  18. anonymous
    • 5 years ago
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    ugh...sub 3x for u and then use the half angle formula to get cosine and sine together, integrate

  19. anonymous
    • 5 years ago
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    sstar are you writing it out? caz i don't really want to...

  20. anonymous
    • 5 years ago
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    >_< nvm, proceed chia, I misunderstood the question lol

  21. anonymous
    • 5 years ago
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    oneprince, did you read what i wrote? is that enough?

  22. anonymous
    • 5 years ago
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    AMISTRE! you can do this! ^_^

  23. amistre64
    • 5 years ago
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    Howdy :)

  24. anonymous
    • 5 years ago
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    lol

  25. anonymous
    • 5 years ago
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    my brain is defunctioning, so I don't think I can write it down even though I took it this semester and it looks SIMPLE! would you amistre? ^_^

  26. amistre64
    • 5 years ago
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    [S] sin^4(3x) dx this is missing something lol this came from something similar to -cos^5(3x) so what do we get when we derive that? and itll tell us what to add...

  27. anonymous
    • 5 years ago
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    it's not really missing something, you can let u = 3x and solve normally like chia said

  28. anonymous
    • 5 years ago
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    if only I remember lol >_<

  29. amistre64
    • 5 years ago
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    Dx(-cos^5(3x)) = 5(3) sin^4(3x) which would be easy to solve right? So we need a "15" in front of your integral to convert it

  30. amistre64
    • 5 years ago
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    lets multiply your original problem by "1"; or rather a convenient for of "1" such as 15/15..

  31. anonymous
    • 5 years ago
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    where did you get cos^5?

  32. amistre64
    • 5 years ago
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    then we can pull out the bottom "15" and integrate the rest up like normal

  33. amistre64
    • 5 years ago
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    sin^4 comes from cos^5; or at least a version of it right?

  34. anonymous
    • 5 years ago
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    you can do this : \[\int\limits (1-\cos^2(3x))^2 dx = \int\limits(1-2\cos^2u+\cos^2u)du\] and simply solve, right?

  35. anonymous
    • 5 years ago
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    where u = 3x ? :)

  36. anonymous
    • 5 years ago
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    right? .-.

  37. anonymous
    • 5 years ago
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    isn't it simpler that way, lol?

  38. amistre64
    • 5 years ago
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    \[\int\limits_{} \frac{5(3)}{5(3)} \sin^4 (3x) dx\]

  39. anonymous
    • 5 years ago
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    why?

  40. anonymous
    • 5 years ago
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    amistre, i think you're doing it wrong.. and i feel bad caz i'm not actually doing it...

  41. amistre64
    • 5 years ago
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    \[\frac{1}{15} \int\limits_{} 5(3) \sin^4 (3x) dx\]

  42. amistre64
    • 5 years ago
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    -cos^5 (3x)/15 + C

  43. anonymous
    • 5 years ago
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    use the half angle formula to integrate sin or cos with a power you MUST have sin * cos

  44. anonymous
    • 5 years ago
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    no, amistre, that's definitely wrong T-T

  45. anonymous
    • 5 years ago
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    Why don't you make a substitution of \[u=3x \rightarrow dx = \frac{du}{3}\]so that then\[I=\frac{1}{3}\int\limits_{}{}\sin^4 u du\]and now use a reduction formula?

  46. amistre64
    • 5 years ago
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    derive it back again, its right.... it should be right lol

  47. anonymous
    • 5 years ago
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    using the way I cut it down, you'll get : \[= \int\limits1 du - 2\int\limits \cos^2u du + \int\limits \cos^2u du\] then you'll use for cos^2 u = 1/2(1+cos2u)

  48. anonymous
    • 5 years ago
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    star is right

  49. anonymous
    • 5 years ago
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    ^_^ there, I hope it's right

  50. anonymous
    • 5 years ago
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    as simple as that :)

  51. anonymous
    • 5 years ago
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    sstar is completely right though i have a question where's the person who ASKED this question

  52. anonymous
    • 5 years ago
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    LOL! watching us, hey oneprince did you understand what we did? ^_^

  53. anonymous
    • 5 years ago
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    wait...he left ._.

  54. anonymous
    • 5 years ago
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    yeah -_- but you got it sstar i have integrating cosine sins T-T so messy...

  55. anonymous
    • 5 years ago
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    \[I=\frac{1}{3}(\frac{1}{32} (12 u-8 \sin(2 u)+\sin(4 u))+c)\]where u = 3x.

  56. anonymous
    • 5 years ago
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    o_o....no wait O_O!

  57. anonymous
    • 5 years ago
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    that's the final answer?

  58. anonymous
    • 5 years ago
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    Sub. u=3x, then du = 3dx --> dx = du/3. Then you have 1/3*int[(sin^4(u) du]

  59. anonymous
    • 5 years ago
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    You can use a reduction formula on sin^4(u)...

  60. anonymous
    • 5 years ago
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    no need for reduction formula ^_^ + i got you lol :)

  61. amistre64
    • 5 years ago
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    what happens when we derive:\[- \frac{\cos^5 (3x)}{15} + C\] ??

  62. anonymous
    • 5 years ago
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    why do you want to derive it? that's not the answer lol? ._.

  63. anonymous
    • 5 years ago
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    He wants to see if it matches the integrand.

  64. anonymous
    • 5 years ago
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    oh

  65. amistre64
    • 5 years ago
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    lol.... if its not the answer, then what does it derive to?

  66. anonymous
    • 5 years ago
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    there's only one way to find out, T.R.Y ^_^

  67. anonymous
    • 5 years ago
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    give me a sec, i'll do it

  68. anonymous
    • 5 years ago
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    I won't even try since my page is lagging BADLY

  69. anonymous
    • 5 years ago
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    anyhow, I'm off to review for my DM test tomorrow, good luck all ^_^

  70. anonymous
    • 5 years ago
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    225 cos(3x)*sin(3x)^4

  71. anonymous
    • 5 years ago
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    Omg, I gave you the answer ^^

  72. anonymous
    • 5 years ago
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    wait i lied! i did * 15 instead of / 15 give me another min :P

  73. anonymous
    • 5 years ago
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    what answer ._.

  74. anonymous
    • 5 years ago
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    ohh i see why you put that 15 there it's cos(3x)*sin(3x)^4

  75. anonymous
    • 5 years ago
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    cheers :D

  76. anonymous
    • 5 years ago
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    Aren't you trying to find the integral of sin^4(3x)?

  77. anonymous
    • 5 years ago
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    yeah but amistre is trying to show that he's right but i showed that he's wrong ^^

  78. amistre64
    • 5 years ago
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    lol...not trying to show that im right, just trying to figure out what I did wrong :)

  79. anonymous
    • 5 years ago
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    :x sorry i didn't mean it like that but do you get it now?

  80. amistre64
    • 5 years ago
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    I see where my error cropped up ;)

  81. anonymous
    • 5 years ago
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    lols kewls

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