Integral of csc2 is -ctn

  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. mathman

    mathman 6,566
    Science Advisor
    Gold Member

    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. Re: Integration

    cos*sec^2(x) or cosx*sec^2(x)?
     
  5. Re: Re: Integration

    Isn't it csc(x)?
     
  6. 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
    Thnxs
     
  7. How about:

    [tex]
    \begin{align*}
    \int_0^\pi\theta^2\csc^2\theta\;d\theta
    &=\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 \\
    &=\int_0^1\theta^2\csc^2\theta\;d\theta+\left[-\cot\theta\right]_1^\pi
    \end{align*}
    [/tex]

    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. master_coda:
    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.

    himanshu121:
    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)).

    jk
     
    Last edited: Dec 11, 2003
  9. 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)



    [tex]

    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)
    [/tex]

    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. Hurkyl

    Hurkyl 16,089
    Staff Emeritus
    Science Advisor
    Gold Member

    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. chroot

    chroot 10,426
    Staff Emeritus
    Science Advisor
    Gold Member

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

    - Warren
     
  12. 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. 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.

    jk
     
    Last edited: Dec 13, 2003
  14. 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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