I was refreshing multiple integral, and I came across with this question:
$\int _0^9\int _{x^{1/2}}^3\cos{(y^3-3y)}\,dy\,dx$
I used change of order since $\cos(y^3-3y)$ cannot be integrated at the current order. I found:
$\int _0^3\int _{0}^{y^2}\cos{(y^3-3y)}\,dx\,dy$
I remember doing a question similar to this but then it was different since when you change order of integration, the lowest value was not 0. It was 1. Had the integration starts from 1 u-substitution would be applicable. But now it is hard.
I have:
${u = y^3 - 3y}$
${du/dy = 3(y^2 - 1)}$
$\int _0^3{y^2}\sin({y^3-3y})dy$
Since, the range was ${0 <= x <= y^2 }$, the zero leaves us with one integral. I know how to work with 1 instead of 0, where I will have:
$\int _0^3{y^2}\sin({y^3-3y})\,dy$ - $\int _0^3\sin({y^3-3y})\,dy$
and factoring:
$\int _0^3{(y^2-1)}\sin({y^3-3y})\,dy$
In the immediate above case, we can just do a simple $u$-substitution.
I don't know how to proceed with when it is zero.