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# Integral with a trick to it watch

1. Argh, I had a question all figured out, and was pleased with the little trick I'd done to work it out, and I didn't write it down. Now I've forgotten it!

Integral between -pi and pi of (x*sinx)

All I wrote down was it was equal to 0. There was some way I rearranged it, or related it to, a simpler integral.... or something.

Any help appreciated.

DMST
2. It's 2pi though.
If it was xcosx or just sinx from -pi to pi it would be 0 because those are odd functions.
3. Is it? The integral of xsinx = sinx - xcosx
[sin(pi) - pi*cos(pi)] - [sin(-pi) - (-pi)*cos(-pi)] = pi - pi = 0

Or am I wrong somewhere?

BLEH. What I meant was the following:

[sin(pi) - pi*cos(pi)] - [sin(-pi) - (-pi)*cos(-pi)] = ( 0 - (-pi)) - ( 0 - pi) = pi - -pi = 2pi
4. Integrating by parts:

u = x v'=sinx
u' = 1 v=-cosx

Hence:

I = [-x.cos(x) + sin(x)]

So

I = {-pi.cos(pi) + sin(pi)} - {pi.cos(-pi) + sin(pi)}
= (pi + 0) - (-pi + 0)
= 2pi as said above

I hope that's correct (and clear).

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Updated: October 18, 2006
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