Having problems solving a double integral

Question

I have to solve the integral below, and I know the answer is zero.

$$\int_{-14}^{14} \int_{392}^{1081} e^{-(x^2+9y^2)}\sin(y^{191})\,dy\,dx$$

I believe I should split the integrals into $\int_{-14}^{14} \sin(y^{191})\,dy$ and $\int_{392}^{1081} e^{-(x^2+9y^2)}\,dx$ and do the product of both. $\int_{-14}^{14} \sin(y^{191})\,dy$ is zero because $\sin$ is an odd function and the interval is symmetric, but can I split the integrals that way? Shouldn't the new integrals be a function of only one variable?

$$\int_{392}^{1081} e^{-x^2}\,dx\;\;\int_{-14}^{14} \sin(y^{191})e^{-9y^2}\,dy$$

But then I can't think of a way to solve the integral above.

Answer 1

Hint: $$\sin(y^{191})e^{-9y^2}=-\sin((-y)^{191})e^{-9(-y)^2}.$$

Answer 2

In order to split. The integrals and write as a product of two integrals you have to be able to separate the variables.

$\int_{392}^{1081} e^{-x^2}\,dx$$\int_{-14}^{14} sin(y^{191})e^{-9y^2}\,dy$ is correct

Now the second integral is integral for odd function so it’s zero.