# Integral of csc2 is -ctn

1. Dec 10, 2003

### 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

Last edited: Dec 10, 2003
2. Dec 10, 2003

### 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.

3. Dec 10, 2003

### PrudensOptimus

Re: Integration

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

4. Dec 10, 2003

### master_coda

Re: Re: Integration

Isn't it csc(x)?

5. Dec 10, 2003

### himanshu121

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

6. Dec 10, 2003

### master_coda

\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*}

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.

7. Dec 10, 2003

### 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

Last edited: Dec 11, 2003
8. Dec 11, 2003

### himanshu121

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)

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

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

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$$\int_0^\pi \cot\theta d\theta = \log(csc\theta-cot\theta)$$

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
9. Dec 11, 2003

### Hurkyl

Staff Emeritus
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. )

10. Dec 11, 2003

### chroot

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

- Warren

11. Dec 11, 2003

### 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, $\csc\theta-\cot\theta=\frac{1-\cos\theta}{\sin\theta}$.

12. Dec 13, 2003

### 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

Last edited: Dec 13, 2003
13. Dec 17, 2003

### harsh

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