# Reduction formula problem

• Jan 31st 2011, 03:08 PM
Stonehambey
Reduction formula problem
Hi,

Here is the problem. The integral $I_n$ is defined by

$I_n = \int_0^{\pi}(\pi / 2 - x)\sin(nx + x/2)\mbox{cosec}(x/2) \, \mbox{d}x$

where $n$ is a positive integer. Evaluate $I_n - I_{n-1}$, and hence evaluate $I_n$ leaving your answer in the form of a sum.

I was hoping someone could tell me if I was on the right track or not. Firstly, is this an improper integral? If it is, how would I start the problem.

If it's not, I did the following

$I_n - I_{n-1} = \int^{\pi}_0 (\pi/2 - x)\mbox{cosec}(x/2)(\sin(nx + x/2) - \sin(nx - x/2)) \,\mbox{d}x$

$= \int^{\pi}_0 (\pi/2 - x)\mbox{cosec}(x/2)(2\cos(nx)\sin(x/2)) \, \mbox{d}x$

I fiddled about with this for a while and managed to get something but of course it's meaningless if the first part is wrong.

Any help would be much appreciated. :)

Thanks

Stonehambey
• Jan 31st 2011, 07:41 PM
tonio
Quote:

Originally Posted by Stonehambey
Hi,

Here is the problem. The integral $I_n$ is defined by

$I_n = \int_0^{\pi}(\pi / 2 - x)\sin(nx + x/2)\mbox{cosec}(x/2) \, \mbox{d}x$

where $n$ is a positive integer. Evaluate $I_n - I_{n-1}$, and hence evaluate $I_n$ leaving your answer in the form of a sum.

I was hoping someone could tell me if I was on the right track or not. Firstly, is this an improper integral? If it is, how would I start the problem.

If it's not, I did the following

$I_n - I_{n-1} = \int^{\pi}_0 (\pi/2 - x)\mbox{cosec}(x/2)(\sin(nx + x/2) - \sin(nx - x/2)) \,\mbox{d}x$

$= \int^{\pi}_0 (\pi/2 - x)\mbox{cosec}(x/2)(2\cos(nx)\sin(x/2)) \, \mbox{d}x$

I fiddled about with this for a while and managed to get something but of course it's meaningless if the first part is wrong.

Any help would be much appreciated. :)

Thanks

Stonehambey

Yes, the integral is improper since $\csc x$ isn't defined at zero, but what you did is fine, imho:

$\displaystyle{I_n-I_{n-1}=\int\limits^\pi_0 \left(\frac{\pi}{2} - x\right)\csc\frac{x}{2}\cdot 2\cos nx\cdot\sin\frac{x}{2}\,dx=2\int\limits^\pi_0\left (\frac{\pi}{2}-x\right)\cos nx\,dx=}$

$\displaystyle{=\frac{\pi}{n}\sin nx|\limits^\pi_0-2\int\limits^\pi_0 x\cos nx\,dx=\frac{-2}{n^2}\cos nx|\limits^\pi_0=\frac{2}{n^2}(1-(-1)^n)=\left\{\begin{array}{ll}0&,\,\mbox{if n is even}\\{}\\\frac{4}{n^2}&,\,\mbox{if n is odd\end{array}\right.}$

...or I'm wrong. Check carefully the above.

Tonio
• Feb 1st 2011, 12:28 AM
Stonehambey
Thanks for the reply :)

Quote:

Originally Posted by tonio

$\displaystyle{=\frac{\pi}{n}\sin nx|\limits^\pi_0-2\int\limits^\pi_0 x\cos nx\,dx=\frac{-2}{n^2}\cos nx|\limits^\pi_0=\frac{2}{n^2}(1-(-1)^n)=\left\{\begin{array}{ll}0&,\,\mbox{if n is even}\\{}\\\frac{4}{n^2}&,\,\mbox{if n is odd\end{array}\right.}$

...or I'm wrong. Check carefully the above.

Tonio

This was exactly the result I got before I posted the question, which is somewhat comforting. So the two questions I have remaining would be

1) How does the fact that it's an improper integral change the result? (I did all the above working before I noticed this, d'oh!)

2) What is the question hinting at when it asks $I_n$ to be written in the form of a sum?

Stonehambey
• Feb 1st 2011, 03:23 AM
tonio
Quote:

Originally Posted by Stonehambey
Thanks for the reply :)

This was exactly the result I got before I posted the question, which is somewhat comforting. So the two questions I have remaining would be

1) How does the fact that it's an improper integral change the result? (I did all the above working before I noticed this, d'oh!)

I supose you must check the trigonometric equality inside the integral sign, as this makes the problematic factor

$\csc(x/2)$ dissapear. Nevertheless I think it was correctly done and thus you changed an improper integral

into a proper one.

2) What is the question hinting at when it asks $I_n$ to be written in the form of a sum?

This one baffles me. This problem ressembles some calculations related to Fourier series of some function, so I

guess it could be possible that what you just evaluated are the coefficients of the cosine part of the Fourier series and

thus you must use this. I can't say for sure.

Tonio

Stonehambey

.
• Feb 1st 2011, 05:15 AM
Stonehambey
The question was lifted from the 1998 STEP II paper (Cambridge entrance exams) and since Fourier series are not on the A-level syllabus I don't think you need to use them in your answer, even if the problem is related to them (This is not uncommon).

Thanks for the help :)

Stonehambey
• Feb 1st 2011, 09:21 AM
tonio
Quote:

Originally Posted by Stonehambey
The question was lifted from the 1998 STEP II paper (Cambridge entrance exams) and since Fourier series are not on the A-level syllabus I don't think you need to use them in your answer, even if the problem is related to them (This is not uncommon).

Thanks for the help :)

Stonehambey

Oh, I think I got it! I just forgot that what I (we) calculated is $I_n-I_{n-1}$ , and not $I_n$ itself, so we got a recursive formula:

$\displaystyle{I_n-I_{n-1}=\left\{\begin{array}{ll}0&,\,\mbox{if n is even}\\{}\\\frac{4}{n^2}&,\,\mbox{if n is odd\end{array}\right.}$ , so we can try:

$I_1=I_1$

$I_2-I_1=0$

$\displaystyle{I_3-I_2=\frac{4}{9}}$

.........................................

$\displaystyle{I_n-I_{n-1}=\frac{4}{n^2}}$ . (** Odd n. For even n take the preceeding value of n)

And now add both columns above:

$\displaystyle{I_n=I_1+4\sum\limits^n_{k=2}\frac{1} {(2n-1)^2}}$ . (Read ** above)

Tonio