cosh(x^2).dx.dy for 3y < or equal to: x < or equal to 3

and 0 < or equal to: y < or equal to 1

I changed the order of integration to get:

cosh(x^2).dy.dx for x/3 < or equal to: y < or equal to 1

and 0 < or equal to: x < or equal to 3

From here I used the fact that cosh(x^2) = 1/2e^(x^2) + 1/2e^(-x^2)

So I have:

1/2e^(x^2) + 1/2e^(-x^2).dy.dx

for x/3 < or equal to: y < or equal to 1

and 0 < or equal to: x < or equal to 3

From here I integrated and ended up with an answer of 0.

I'm not sure if the approach I used was correct. can i use cosh(x^2) = 1/2e^(x^2) + 1/2e^(-x^2) for the double integration.

Any advice on whether I went about this the right way would be greatly appreciated!