Sylva Posted October 12, 2017 Share Posted October 12, 2017 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. Link to comment Share on other sites More sharing options...
Country Boy Posted October 13, 2017 Share Posted October 13, 2017 (edited) 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 Link to comment Share on other sites More sharing options...
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