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marigirl

  • one year ago

Integration question:

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  1. marigirl
    • one year ago
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    \[-cosec(6x)\cot(6x)\]

  2. marigirl
    • one year ago
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    \[-cosec(6x)\cot(6x)\] \[\frac{ -1 }{ \sin(6x) }*\frac{ \cos (6x) }{\sin (6x)}\] Integrate \[\frac{ -\cos(6x) }{ \sin^2(6x) }\]

  3. misty1212
    • one year ago
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    HI!!

  4. marigirl
    • one year ago
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    Hi Misty1212 I need guidance from here plzz

  5. misty1212
    • one year ago
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    do you know a function whose derivative is \[\csc(x)\cot(x)\]?

  6. misty1212
    • one year ago
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    hint, you should !

  7. misty1212
    • one year ago
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    if you do not, i can tell you, but i assure you it is in your book under the section of "derivatives of trig functions"

  8. marigirl
    • one year ago
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    if y=cosec(x) then y'=-cosec(x)cot(x)

  9. misty1212
    • one year ago
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    ok you got it

  10. misty1212
    • one year ago
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    of course you have to adjust for the \(6x\) but that is a mental u - sub divide by 6 to take care of that

  11. misty1212
    • one year ago
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    it it clear what i mean?

  12. misty1212
    • one year ago
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    you need to change the sign, and divide by 6 to get the derivative you want, i.e. \[-\frac{1}{6}\csc(6x)\] will work

  13. marigirl
    • one year ago
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    yes because when you integrate you tend to divide by the power multiplied by the coefficient of x?

  14. misty1212
    • one year ago
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    yes it is the chain rule backwards

  15. misty1212
    • one year ago
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    the check is easy right? take the derivative of \[-\frac{1}{6}\csc(6x)\] and see that you get the integrand \[\csc(6x)\cot(66x)\]

  16. misty1212
    • one year ago
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    oops too many sixes, but you get the idea

  17. marigirl
    • one year ago
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    this is a once step thing which is fantastic thank you! but if for some reason I decided to make things complicated and go \[\frac{ -\cos(6x) }{ \sin^2(x) }\] can i use \[\int\limits \frac{ f'(x) }{ f(x) } \] rule somehow?

  18. misty1212
    • one year ago
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    you still need the mental u - sub

  19. misty1212
    • one year ago
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    and it would not be what you wrote, because of the square in the denominator

  20. misty1212
    • one year ago
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    it would be more like \[\int\frac{f'(x)}{f^2(x)}dx\]

  21. marigirl
    • one year ago
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    sorry i meant \[\frac{ -\cos(6x) }{ \sin^2(6x) }\]

  22. marigirl
    • one year ago
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    Sorry I didnt think I made myself very clear Could I possibly use \[A \int\limits \frac{ f'(x) }{ f(x) } dx= A \ln \left| fx \right| +c\]

  23. misty1212
    • one year ago
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    no, as i said, you are missing the square

  24. misty1212
    • one year ago
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    \[\int\frac{f'(x)}{f^2(x)}dx\]j you will not get the log

  25. marigirl
    • one year ago
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    oh yea i get it now .. best method is to use the basic rules outlined . Thanks Heaps! I cant believe i made it so complicated lol

  26. misty1212
    • one year ago
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    lol it is easy to do with integration look for the simple first \[\color\magenta\heartsuit\]

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