Can the Integral of csc2 be Simplified Using the Tabular Method?

[himanshu121]

Here is the Problem:

\int_0^\pi\theta^2cosec^2\theta d\theta


I want to do this problem by parts and finding hard to find shortest way to do the problem any shortest way pls

 

[mathman]

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.

 

[PrudensOptimus]

Originally posted by himanshu121
Here is the Problem:

\int_0^\pi\theta^2cosec^2\theta d\theta


I want to do this problem by parts and finding hard to find shortest way to do the problem any shortest way pls

cos*sec^2(x) or cosx*sec^2(x)?

 

[master_coda]

Originally posted by PrudensOptimus
cos*sec^2(x) or cosx*sec^2(x)?

Isn't it csc(x)?

 

[himanshu121]

It is cosecant(x) i.e csc(x)

Integrating log(sin) may take a little work

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

 

[master_coda]

How about:

<br /> \begin{align*}<br /> \int_0^\pi\theta^2\csc^2\theta\;d\theta<br /> &amp;=\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\theta^2\csc^2\theta\;d\theta \\<br /> &amp;&gt;\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\csc^2\theta\;d\theta \\<br /> &amp;=\int_0^1\theta^2\csc^2\theta\;d\theta+\left[-\cot\theta\right]_1^\pi<br /> \end{align*}<br />

Now since \lim\limits_{\theta\rightarrow\pi}(-\cot\theta)=+\infty 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.

 

[jk7711]

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

 

[himanshu121]

Best way to do it would be to plug it into a calculator or look it up in a table

We are not allowed to use calculator in India till we are undergraduate
I found the way but don't know whether it is shortest one or not but definately i won't stuck at log(sinx)



<br /> <br /> I = \int_0^\pi\theta^2\csc^2\theta d\theta

I= \int_0^\pi\ (\pi-\theta)^2\csc^2\theta d\theta

this gives

\pi\int_0^\pi\csc^2 \theta d\theta = 2\int_0^\pi\theta\csc^2\theta d\theta

integrating by parts with one part as \theta and other as \theta\csc^2\theta d\theta

i will get\int_0^\pi \cot\theta d\theta = \log(csc\theta-cot\theta)<br />

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

 

[Hurkyl]

I agree with master_coda...

the integral of x^2*(cscx)^2 from 1 to pi comes out to be about 4.2

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. )

 

[chroot]

Originally posted by Hurkyl
(though, 3*10^14 is a good approximation of infinity. )

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

- Warren

 

[master_coda]

I'm pretty sure that the integral diverges. My algebra, my calculator and my computer all agree.

Also, if it's any help, \csc\theta-\cot\theta=\frac{1-\cos\theta}{\sin\theta}.

 

[jk7711]

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

 

[harsh]

Why don't 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