Define for all $n\in\mathbb{N}$, a definite double integral, $$I_n= \int_{0}^{1}\int_{0}^{1}\left(\frac{x^2(1-x)y^2(1-y)}{1+x^2 y^2}\right)^n\frac{\cos((2n+1)\tan^{-1}(xy))}{\sqrt{1+x^2y^2}}\ dxdy$$
I need asymptotic formula of $I_n$ as $n\to\infty$
Firstly we see that since $x(1-x)\leq \frac{1}{4}$, so the definite integral is bounded since $$|I_n|\leq \frac{1}{16^n}$$ I tried using Laplace's method by first substituting $u=2x-1$ and $v=2y-1$ so that $x=\frac{u+1}{2}$, $y=\frac{v+1}{2}$ and $dx=\frac{du}{2}$, $dy=\frac{dv}{2}$. Both the limit of integration becomes $-1$ to $1$.
$$I_n=\frac{1}{4}\int_{-1}^{1}\int_{-1}^{1} \left(\frac{\left(\frac{u+1}{2}\right)^2\left(\frac{v+1}{2}\right)^2\frac{(1-u)(1-v)}{4}}{1+\frac{(u+1)^2(v+1)^2}{16}}\right)^n \frac{\cos((4n+1)\tan^{-1}\left(\frac{(u+1)(v+1))}{4}\right) }{\sqrt{1+\frac{(u+1)^2(v+1)^2}{16}}}dxdy$$ So we obtain $$I_n=\frac{1}{4}\int_{-1}^{1}\int_{-1}^{1} \frac{e^{n\log f(u,v)}\cos((4n+1)\tan^{-1}\left(\frac{(u+1)(v+1))}{4}\right)}{\sqrt{1+\frac{(u+1)^2(v+1)^2}{16}}}dxdy$$ where $$f(u,v)=\frac{\left(\frac{u+1}{2}\right)^2\left(\frac{v+1}{2}\right)^2\frac{(1-u)(1-v)}{4}}{1+\frac{(u+1)^2(v+1)^2}{16}}$$ Any help would be highly appreciated. Thank you.
Edit 1 Can we use some result say for instance pg. 331 from the book Asymptotic expansions of integrals by N Bleistein, R.A. Handelsman, 2010?
Edit 2 I am also curious about will such an asymptotic formula exist for this definite double integral? If yes, then say $I_n\sim a_n$ as $n\to \infty$. Then is $a_n\neq 0$ $\forall n\in\mathbb{N}$?