Here's the question you clicked on:
tomiko
need help with this double integral question :) http://gyazo.com/a3e6e7133a02d92949f4a14f0cb3961e
What are you stuck with?
\[\int\limits_{\pi}^{2\pi}\int\limits_{0}^{1}xsin(xy)dxdy\]
since your limits are constants; you can prolly more easily swap it for dydx
which proff did ... ironically :)
IRONICALLY?!?! LOOOOOOOOOOL
\[\int\limits_{\pi}^{2\pi}\int\limits_{0}^{1}xsin(xy)~dydx\] \[\int\limits_{\pi}^{2\pi}\left(\int\limits_{0}^{1}xsin(xy)~dy\right)dx\]
doh ... loathesome integrals!!
@amistre64 doesn't the zero to one have to stick with the dx differential though?
\[\int_{x=0}^{x=1}\left(\int_{y=\pi}^{y=2\pi}xsin(xy)~dy\right)dx\]
it tends to help me out if i label the integrals ....
Isn't the general integral of that xsin(xy)/y|
\[\int_{x=0}^{x=1}\left(\int_{y=\pi}^{y=2\pi}xsin(xy)~dy\right)dx\] \[\int_{x=0}^{x=1}-cos(2\pi~x)+cos(\pi~x)~dx\]
I thought you didn't have to do parts integration since your X term out front evidently doesn't have a Y in it?
you still have to integrate your limits respectively x = [0,1] has to address your dx y = [pi, 2pi] has to address your dy
the dx, dy parts tells you what you focus on while leaving the other term a constant
LOLOL I forgot about that
I didn't realize you were evaluating limits in your 2nd step of that response
let x = k int k sin(ky) dy -> -cos(ky) -cos(x 2pi) - -cos(x pi)
int cos(x pi) - cos(x 2pi) dx sin(x pi)/pi - sin(x 2pi)/2pi (sin(pi)/pi - sin(2pi)/2pi) - (sin(0 pi)/pi - sin(0 2pi)/2pi) (0 - 0) - (0 - 0) so with any luck :)