The function I am trying to integrate is:
$$\int_0^{12}\int_{y/4}^3 \sin(x^2)\ \mathrm{d}y\ \mathrm{d}x$$
I am having trouble finding out where to start. I have tried to just go ahead and integrate, but then I would have to deal with calculating cosine of $9$ and $\left(\frac{y}{4}\right)^3$. I thought about changing the order of integration, but how would you do that considering the inner bound is a function of $x$?
I would greatly appreciate some help to point me in the right direction.
$$ \begin{align} I&=\int_0^{12}\left( \int_{y/4}^{3}\sin (x^2)dx\right)dy\\&=\int_0^3dx\int_0^{4x} \sin(x^2)dy\\&=\int_0^3\sin(x^2)4xdx \\ &= 2\int_0^3\sin(x^2)d(x^2)\\&=2\int_0^9\sin udu \\&=2(1-\cos 9) \end{align} $$