Evaluate double integral $$\iint x^2 \cos(x^2-xy)\,dx\,dy$$ where region is bounded by sides of triangle whose vertices are $(0,0),(1,0),(0,1)$.
I used order $dy\,dx$ to evaluate it. I becomes unsolvable after this. Where i am wrong?
It reduces to $$\int_{0}^{1} -x\sin(2x^2-x) + x\sin(x^2)$$
EDIT
As per request of user Jester Tran
My double integral was $$\int_{0}^{1} \int_{0}^{1-x} x^2 \cos(x^2-xy)\,dx\,dy$$
try solving this integral $$\int_{o}^{1}\int_{0}^{1-x}x^2\cos({x^2-xy})dy dx$$ your final fom contains $\cos$ where as it should be $sin$ as far as solving the integral goes you can take the term inside $\sin$ as $t$ for 2 different integrals solution : $$\int_{0}^{1}-x\sin{(2x^2-x)+\int_{0}^{1}x\sin{x^2}}=\int_{0}^{1}-(x-1/4)\sin{(2x^2-x)}+\int_{0}^{1}x\sin{x^2}-\int_{0}^{1}\frac{1}{4}\sin{(2x^2-x)}$$ last integral does not have an elementary solution