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Integral of csc2 is -ctn

  1. Dec 10, 2003 #1
    Here is the Problem:

    [tex] \int_0^\pi\theta^2cosec^2\theta d\theta[/tex]

    I want to do this problem by parts and finding hard to find shortest way to do the problem any shortest way pls
    Last edited: Dec 10, 2003
  2. jcsd
  3. Dec 10, 2003 #2


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    The straightforward way is probably the easiest. integral of csc2 is -ctn and integral of ctn is log(sin). Integrating log(sin) may take a little work.
  4. Dec 10, 2003 #3
    Re: Integration

    cos*sec^2(x) or cosx*sec^2(x)?
  5. Dec 10, 2003 #4
    Re: Re: Integration

    Isn't it csc(x)?
  6. Dec 10, 2003 #5
    It is cosecant(x) i.e csc(x)
    This is probably the easiest way of doing which i too have tried but it is not the shortest way

    There are many ways of doing a problem i am looking for shortest way
  7. Dec 10, 2003 #6
    How about:

    &=\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\theta^2\csc^2\theta\;d\theta \\
    &>\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\csc^2\theta\;d\theta \\

    Now since [itex]\lim\limits_{\theta\rightarrow\pi}(-\cot\theta)=+\infty[/itex] we know that the rightmost term diverges. Moreover, the left integral (the one from 0 to 1) is clearly positive. Thus the original integral clearly diverges.
  8. Dec 10, 2003 #7
    the integral of x^2*(cscx)^2 from 1 to pi comes out to be about 4.2
    the integral of (cscx)^2 from 1 to pi goes to inifinite. I believe to say the integral of x^2*(cscx)^2 from 1 to pi is greater than the integral of (cscx)^2 from 1 to pi since according to the values it came out as then the formula should be switched around and divergence would not be proved.

    Best way to do it would be to plug it into a calculator or look it up in a table. Otherwise you're probably stuck with integrating log(sin(x)).

    Last edited: Dec 11, 2003
  9. Dec 11, 2003 #8
    We are not allowed to use calculator in India till we are undergraduate
    I found the way but dont know whether it is shortest one or not but definetly i wont stuck at log(sinx)


    I = \int_0^\pi\theta^2\csc^2\theta d\theta [/tex]

    [tex]I= \int_0^\pi\ (\pi-\theta)^2\csc^2\theta d\theta [/tex]

    this gives

    [tex] \pi\int_0^\pi\csc^2 \theta d\theta = 2\int_0^\pi\theta\csc^2\theta d\theta [/tex]

    integrating by parts with one part as [tex]\theta[/tex] and other as [tex]\theta\csc^2\theta d\theta [/tex]

    i will get[tex] \int_0^\pi \cot\theta d\theta = \log(csc\theta-cot\theta)

    much easier than integrating log(sinx)

    But another problem is how i will put the limits in cotx from 0 to pi in both cases it is infinity and i know there is no break in the function cotx b/w these points
    Last edited: Dec 11, 2003
  10. Dec 11, 2003 #9


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    I agree with master_coda...

    Where did you come up with this?! My calculator says 3*10^14 with the caveat of "questionable accuracy" (though, 3*10^14 is a good approximation of infinity. :wink:)
  11. Dec 11, 2003 #10


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    Yes, for large values of 3*10^14.

    - Warren
  12. Dec 11, 2003 #11
    I'm pretty sure that the integral diverges. My algebra, my calculator and my computer all agree.

    Also, if it's any help, [itex]\csc\theta-\cot\theta=\frac{1-\cos\theta}{\sin\theta}[/itex].
  13. Dec 13, 2003 #12
    hurkyl. yeah that 4.2 was kinda off huh? just keeping you on your toes i guess:P i entered it in wrong but i redid it and came up with what you got.

    i used u=(x^2)csc^2(x) and dv=dx and came up with the first term (x^3*csc^2(x)) going to infinite so i think you might be right coda.

    Last edited: Dec 13, 2003
  14. Dec 17, 2003 #13
    Why dont you try using the tabular method to doing this problem.
    Since theta^2 will eventually go to zero if you keep on taking the derivatives, you should do it by tabular method
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