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angela210793

  • 3 years ago

@satellite73

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  1. angela210793
    • 3 years ago
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    |dw:1341334653707:dw|

  2. satellite73
    • 3 years ago
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    ick i think the idea is to write the denominator as a single trig function

  3. angela210793
    • 3 years ago
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    :O i thought maybe to split and take 2 fractions O.o

  4. angela210793
    • 3 years ago
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    but idk how to do tht -_-

  5. satellite73
    • 3 years ago
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    it is going to be \(5\sin(x+\theta)\) but i don't see a nice form for \(\theta=\tan^{-1}(\frac{4}{3})\)

  6. satellite73
    • 3 years ago
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    actually it doesn't matter because \(\theta\) is a constant also i made a mistake, it is \(5\sin(x-\theta)\) still no matter

  7. satellite73
    • 3 years ago
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    i don't think there is a snappier way to do this. let me think for a second

  8. angela210793
    • 3 years ago
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    okok

  9. satellite73
    • 3 years ago
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    ok no i can't think of a better way, and also i tried wolfram and what a disaster what you need to know is \[a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\] where \(tan(\theta)=\frac{b}{a}\)

  10. angela210793
    • 3 years ago
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    hmmmm....suppose i know tht....how to use it O.o

  11. satellite73
    • 3 years ago
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    well then it is easy, especially if you have a table of integrals this becomes \[\frac{1}{5}\int\csc\left(x-\tan^{-1}(\frac{4}{3})\right)\]

  12. satellite73
    • 3 years ago
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    don't worry about the arctan part, that is a number

  13. satellite73
    • 3 years ago
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    so all you need to do is look up the anti derivative of cosecant and you are done

  14. satellite73
    • 3 years ago
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    here is a video explanation http://www.youtube.com/watch?v=STUh3ni4l50

  15. angela210793
    • 3 years ago
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    hmmmm....we've never use secant...wht is it?

  16. satellite73
    • 3 years ago
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    cosecant it is the reciprocal of sine

  17. satellite73
    • 3 years ago
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    integral is \[\int csc(x)dx=-\ln(\cot(x)+\csc(x))\] if you have not used this i really have no idea how you are supposed to do this problem maybe multiply top and bottom by the conjugate?

  18. angela210793
    • 3 years ago
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    :/....ok.. thank you Sir :D

  19. satellite73
    • 3 years ago
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    maybe we can get some help i will repost there might be a snappy trick

  20. angela210793
    • 3 years ago
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    okk

  21. satellite73
    • 3 years ago
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    maybe myininaya has a snappy way multiply top and bottom by \(3\sin(x)+4\cos(x)\) maybe?

  22. myininaya
    • 3 years ago
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    I don't know @satellite73 I like what you did.

  23. satellite73
    • 3 years ago
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    ok then i will stick to that. thanks

  24. myininaya
    • 3 years ago
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    I actually think what sat did was probably the most snappiest thing you can do for this type of integral

  25. waterineyes
    • 3 years ago
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    If we replace sinx by 2tan(x/2)/(1 + tan^2(x/2),

  26. angela210793
    • 3 years ago
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    sec=1/cosx????

  27. myininaya
    • 3 years ago
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    yes @angela210793 sec(x)=1/(cos(x)) csc(x)=1/(sin(x))

  28. angela210793
    • 3 years ago
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    ok....thanks guys :D

  29. waterineyes
    • 3 years ago
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    \[\huge sinx = \frac{2\tan \frac{x}{2}}{1+ \tan^2\frac{x}{2}}\] \[\huge cosx = \frac{1 - \tan^2\frac{x}{2}}{1+ \tan^2\frac{x}{2}}\]

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