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jaersyn

  • 3 years ago

integrate (cscx)^3 >_>

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  1. RadEn
    • 3 years ago
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    integration by parts it will work

  2. RadEn
    • 3 years ago
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    int (cscx)^3 = int (cscx)^2 cscx dx let u=cscx du=.... dv=(cscx)^2 dx v=int(cscx)^2 dx v=..... and next use parts rule : int(udv) = uv-int(vdu)

  3. RadEn
    • 3 years ago
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    but i was stuck to find int(cscx) dx, @TuringTest ^^ lol

  4. RadEn
    • 3 years ago
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    i'm not sure int (cscx)dx = ln(sinx)

  5. Directrix
    • 3 years ago
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    @RadEn OpenStudy values the Learning process - not the ‘Give you an answer’ process •Don’t post only answers - guide the asker to a solution. http://openstudy.com/code-of-conduct

  6. experimentX
    • 3 years ago
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    \[ \int \csc ^3xdx = \int (1 + \cot^2 x) \csc x \; dx\]

  7. jaersyn
    • 3 years ago
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    is there another trig identity i'm supposed to use? u-substitution isn't working

  8. Jemurray3
    • 3 years ago
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    There is something of a trick to this... try \[\int \frac{ \sin(x)}{\sin^2(x) } dx\]

  9. jaersyn
    • 3 years ago
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    i'm not sure how you got that

  10. jaersyn
    • 3 years ago
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    its not (cosx)^2 / (sinx)^3?

  11. Jemurray3
    • 3 years ago
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    I'm talking about integrating cosecant.

  12. experimentX
    • 3 years ago
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    use the same technique as used here http://en.wikipedia.org/wiki/Integral_of_secant_cubed

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