Hi,

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

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

where $\displaystyle n$ is a positive integer. Evaluate $\displaystyle I_n - I_{n-1}$, and hence evaluate $\displaystyle I_n$ 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

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

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

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