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Integrating csc(x)

  1. Nov 21, 2004 #1
    haven't found a way of doing it so far. I have a feeling that it's extremely easy, and i'm missing how to do it somehow :/
    Last edited: Nov 21, 2004
  2. jcsd
  3. Nov 21, 2004 #2
    do you mean cosec(x), the cosecans ??? If so, just use the t=tan(x/2) formula's...

  4. Nov 21, 2004 #3
    I'm not sure what formula you're talking about.

    [tex] \int \csc(x) dx [/tex]

    If you take [tex]t=tan(\frac{x}{2})[/tex],

    you'd get:

    [tex]\frac {d}{dx} \tan(x)= \frac{1}{2}.sec^2(x)[/tex]

    Not sure how to go from there...
    Last edited: Nov 21, 2004
  5. Nov 21, 2004 #4


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    What marlon meant, is the following:

    Substitute [tex]u=tan(\frac{x}{2})[/tex]
    This implies:
    Hence, we have:
  6. Nov 22, 2004 #5
    I'm not sure how you did this equality:

    [tex]\frac{\cos^{2}(\frac{x}{2 })+\sin^{2}(\frac{x}{2})}{2\sin(\frac{x}{2})\cos(\ frac{x}{2})}=\frac{1+tan^{2}(\frac{x}{2})}{2tan(\frac{x}{2})}[/tex]

    Can you please clarify?

    Other than that, it's all clear, thank you very much.
    Last edited: Nov 22, 2004
  7. Nov 22, 2004 #6
    [itex]\int \csc x = \int \csc x \left(\frac{\csc x - \cot x}{\csc x - \cot x}\right) = \int \frac{du}{u} = \ln |csc x - cot x|[/itex]
    (is that what we're talking about?)
    Last edited: Nov 22, 2004
  8. Nov 22, 2004 #7
    yes, nice method :)

    Still i would like someone to explain the question of my last post.
    Thank you :)
  9. Nov 22, 2004 #8
    separate the fractions and simplify
    [itex]\frac{\cos u}{2\sin u}+ \frac{\sin u}{2\cos u} =[/itex]
    [itex]\frac{1}{2\tan u} + \frac{tan u}{2} [/itex]
    get a common denominator and you're done.
  10. Nov 22, 2004 #9


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    Just divide by cos^2(x/2) in both the numerator and the denominator of the LHS of the expression in question and it will drop straight out.
  11. Nov 22, 2004 #10
    Yes, excellent, so know i know i am stupid hehe :)

    Thanks alot for your help guys :)
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