Double integral Help

[Sylva]

Hey,

I've been stuck on this problem for quite some time:

J = ∫0 ->4 ∫ sqrt(x) -> 2   (1 + y^2 * cos(x * sqrt(y))) dydx

The cos (x * sqrt(y)) is the one causing trouble. I can't seem to find a way to integrate this. I even tried to turn it to polar coordinates but nothing seems to work.

What am I doing wrong? Could someone point me in the right direction?

Thanks in advance.

PS: Sorry for my english, it's not my native language.

[Country Boy]

 I thought this had already been answered.  x ranging from 0 to 4, and y, for each x, ranging from sqrt{x} to 2 is the same as y ranging from 0 to 2 and, for each y, x ranging from y^2 to 4.  That is, this integral is the same as J= \int_0^2\int_{y^2}^4 (1+ y^2 cos(x\sqrt{y})dx dy  To do that, first let u= x\sqrt{y} so that du= \sqrt{y}dx  When x= y^2, u= y^2\sqrt{y}= y^{3/2} and when x= 4, u= 4\sqrt{y}= 4y^{1/2}.

The integral becomes (1/sqrt(y))\int_0^2\int_{y^{3/2}}^{4y^{1/2}} 1+ y^2cos(u)du dy

Edited October 13, 2017 by Country Boy