## tomiko 2 years ago need help with this double integral question :) http://gyazo.com/a3e6e7133a02d92949f4a14f0cb3961e

1. tomiko

@amistre64

2. FutureMathProfessor

What are you stuck with?

3. FutureMathProfessor

$\int\limits_{\pi}^{2\pi}\int\limits_{0}^{1}xsin(xy)dxdy$

4. amistre64

since your limits are constants; you can prolly more easily swap it for dydx

5. amistre64

which proff did ... ironically :)

6. FutureMathProfessor

IRONICALLY?!?! LOOOOOOOOOOL

7. amistre64

$\int\limits_{\pi}^{2\pi}\int\limits_{0}^{1}xsin(xy)~dydx$ $\int\limits_{\pi}^{2\pi}\left(\int\limits_{0}^{1}xsin(xy)~dy\right)dx$

8. amistre64

doh ... loathesome integrals!!

9. FutureMathProfessor

@amistre64 doesn't the zero to one have to stick with the dx differential though?

10. amistre64

$\int_{x=0}^{x=1}\left(\int_{y=\pi}^{y=2\pi}xsin(xy)~dy\right)dx$

11. amistre64

it tends to help me out if i label the integrals ....

12. FutureMathProfessor

Isn't the general integral of that xsin(xy)/y|

13. amistre64

$\int_{x=0}^{x=1}\left(\int_{y=\pi}^{y=2\pi}xsin(xy)~dy\right)dx$ $\int_{x=0}^{x=1}-cos(2\pi~x)+cos(\pi~x)~dx$

14. FutureMathProfessor

I thought you didn't have to do parts integration since your X term out front evidently doesn't have a Y in it?

15. amistre64

16. amistre64

the dx, dy parts tells you what you focus on while leaving the other term a constant

17. FutureMathProfessor

18. FutureMathProfessor

I didn't realize you were evaluating limits in your 2nd step of that response

19. amistre64

let x = k int k sin(ky) dy -> -cos(ky) -cos(x 2pi) - -cos(x pi)

20. amistre64

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 :)

21. tomiko

thanks @amistre64