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anonymous

  • 5 years ago

Intergration please solve intergral tanx^3 dx

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  1. amistre64
    • 5 years ago
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    odd degree trig functions; I think its easier to split them up into groups of something...

  2. anonymous
    • 5 years ago
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    Re-write it as ∫tan^2(x)*tan(x) dx ---> ∫(sec^2(x)-1)*tan(x) dx = ∫sec^2(x)tan(x) - tan(x) dx. Let u=tan(x).

  3. anonymous
    • 5 years ago
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    well, it is quite simple actually, just wait a minute and sstarica will show you

  4. anonymous
    • 5 years ago
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    re-write it and you'll get :\[\int\limits_{}^{}sinx^3/cosx^3 dx\] use u substitution for cosx^3 ^_^ give it a try :)

  5. anonymous
    • 5 years ago
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    LOL andy :D

  6. amistre64
    • 5 years ago
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    Here, use this :)

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  7. anonymous
    • 5 years ago
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    oh, and ssarica, gratz on making a Superstar, I guess I cant call you starlet anymore

  8. anonymous
    • 5 years ago
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    lol, call me whatever you want and thank you ^_^, some of the fanning came from you :)

  9. anonymous
    • 5 years ago
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    yup, whole 4 fannings

  10. anonymous
    • 5 years ago
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    amistre! are you encouraging students on using calculators instead of thinking about it?!

  11. amistre64
    • 5 years ago
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    at 101 fans you get to be a "hero"....its a shame really, all that work just to become a sandwich....

  12. anonymous
    • 5 years ago
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    different times, different accounts, but same me

  13. anonymous
    • 5 years ago
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    :) I don't really care much about the fanning process

  14. amistre64
    • 5 years ago
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    the mr proffesor is more of an "exercise" in math than it is a calculator :)

  15. anonymous
    • 5 years ago
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    lol, anyway did you understand it deidre? ^_^

  16. anonymous
    • 5 years ago
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    @sstarica: I'm not sure if your method works, it ends up giving you -1/3 * ∫1/cos^5(x) dx. :/

  17. amistre64
    • 5 years ago
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    2x6=? 12 (green light)

  18. anonymous
    • 5 years ago
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    for reals? let me try

  19. anonymous
    • 5 years ago
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    Whoops, not quite..

  20. anonymous
    • 5 years ago
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    no lol , it works!

  21. anonymous
    • 5 years ago
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    Oh, sorry. Thought you typed something else ;P

  22. anonymous
    • 5 years ago
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    wait, you're right it doesn't work ^^" sorry for the mess again diedre, ignore my answer. Thank you Quantum ^_^

  23. anonymous
    • 5 years ago
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    deidre is away, geting some hotdogs for me and me

  24. anonymous
    • 5 years ago
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    Now let me check if my answer works...xD

  25. anonymous
    • 5 years ago
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    so dont worry, you have some time to clear the mess before he comes back

  26. anonymous
    • 5 years ago
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    \[du = -3x^2sinx^3dx\] which is wrong. you can't substitute it there, my dearest apologies ^^"

  27. anonymous
    • 5 years ago
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    Ah, let me expand my answer to make it clearer: ∫tan(x)sec^2(x) dx - ∫tan(x) dx. In the first integral, let u=tan(x), and in the second it integrates cleanly. :P

  28. anonymous
    • 5 years ago
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    why did you put tanxsec^2x? where did you get it from

  29. anonymous
    • 5 years ago
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    x^3 acts as an angle here

  30. anonymous
    • 5 years ago
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    and not the power of tan :)

  31. anonymous
    • 5 years ago
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    ok, guys, I am about to go to sleep, just going to workout abit and maybe eat something and i am off.

  32. anonymous
    • 5 years ago
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    same, I'm tired

  33. anonymous
    • 5 years ago
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    oh, nice, so we are going to beds together again

  34. anonymous
    • 5 years ago
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    anyway, solve the question and I'm off to bed, sorry ^_^

  35. anonymous
    • 5 years ago
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    I'm going to bed ALONE >_>

  36. anonymous
    • 5 years ago
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    I sayed BEDS with an S, not BED :D LOL, I knew it you have interesting thoughts in your head :)

  37. anonymous
    • 5 years ago
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    I got it from the relation tan^2(x) = sec^2(x) - 1. ∫tan^3(x)dx = ∫tan^2(x)*tan(x) dx = ∫(sec^2(x)-1)*tan(x) dx = ∫tan(x)sec^2(x) - tan(x) dx = ∫tan(x)sec^2(x) dx - ∫tan(x) dx. :P

  38. anonymous
    • 5 years ago
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    nope :)

  39. anonymous
    • 5 years ago
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    I never think of such things angoo =P

  40. anonymous
    • 5 years ago
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    yes but quantum the question is :\[\int\limits_{}^{}tanx^3\] and not : \[\int\limits_{}^{}\tan^3x\]

  41. anonymous
    • 5 years ago
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    hmm... well, I don't too, becouse things like those should be practice once married

  42. anonymous
    • 5 years ago
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    I agree :)

  43. anonymous
    • 5 years ago
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    quantum you copied the question wrong lol

  44. anonymous
    • 5 years ago
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    good, smart girl, not the "new era" type

  45. anonymous
    • 5 years ago
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    nah, I'm old fashioned ^_^ more classy

  46. anonymous
    • 5 years ago
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    wonderful, smart, classy, what else could one want....

  47. anonymous
    • 5 years ago
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    want what? ._.

  48. anonymous
    • 5 years ago
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    Erm...alright, then we'll see when deidre clears it up. :P

  49. anonymous
    • 5 years ago
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    like in, what else would I look for in a friend

  50. anonymous
    • 5 years ago
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    lol , I agree Quantum

  51. anonymous
    • 5 years ago
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    aww, angoo, you're soo sweet ^_^

  52. amistre64
    • 5 years ago
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    im sticking with my safety net....(S) x^2 dx..... :)

  53. anonymous
    • 5 years ago
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    not sweet, just relaxed and not trying to be someone I am not

  54. anonymous
    • 5 years ago
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    lol amistre! and yes you are sweet andy ^_^

  55. anonymous
    • 5 years ago
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    well, thanks, going to the dream world now, I think I will dream about beautiful land with happy little rabbits and dears and bears living together, and I will be ther too, we will play and have fun....

  56. anonymous
    • 5 years ago
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    lala land, lol, I'm off to bed too, good night ^_^

  57. anonymous
    • 5 years ago
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    and you are welcome :)

  58. anonymous
    • 5 years ago
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    night night :)

  59. anonymous
    • 5 years ago
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    sleep tight ~

  60. anonymous
    • 5 years ago
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    :) I was about to say "don't let the bugs bite" but it is getting old, so..

  61. anonymous
    • 5 years ago
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    lol, bye

  62. anonymous
    • 5 years ago
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    bye

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