Yesterday I was interested in the so-called Sophie Germain's Identity, see this section of Wikipedia. I was interested in it about composite numbers. After I was thinking about different consequences of this identity. Playing I prosoposed calculate the easy $$\int_0^1\int_0^1\frac{x^4+4y^4}{x^2-2xy+2y^2}dxdy,$$ and after I was interested in integrals $$\int_0^1\int_0^1 f\left(\frac{x^4+4y^4}{x^2-2xy+2y^2}\right)dxdy,$$ for different functions than $f(z)=z$. For example $f(z)=\log z$, that can be solved in closed-form.
A different example was $f(z)=\cos z$, using a CAS I believe that it can not solve in closed-form.
Question. Calculate in terms of series, special functions (integral functions) or well-known constants $$\int_0^1\int_0^1\cos\left((x+y)^2+y^2\right)dxdy.\tag{1}$$ Many thanks.
I am thinking how get an approximation of $(1)$ using calculus, thus avoid provide me hints to this purpose.
Notation: $C$ and $S$ are Fresnel integral.
let calculate first integrate thus we can write:
$ \int cos((x+y)^2+y)dx $
using $u=x+y \to dx=du$ we can write it like :
$\int(cos(u+y)du$ expand trigonometric function: $\int(cos(y^2)cos(u^2)-sin(y^2)sin(u^2))du=cos(y^2)\int cos(u^2)du-sin(y^2)\int sin(u^2)du$
calculate first and second integrals with $v={u\sqrt2 \over \sqrt \pi}$ .
and finally with a little calculate :
$- \pi ({sin(y^2)S({\sqrt 2 (x+y) \over \sqrt \pi})-cos(y^2)C({\sqrt 2 (x+y) \over \sqrt \pi}))}\over \sqrt 2$
for the first integral .
to calculate the second one we split them and calculate them sepratly so we write : $$ \begin{align} \int sin(y^2)S({\sqrt 2 (x+y) \over \sqrt \pi})dy=sin(y^2)((x+y)S((x+y) \sqrt { 2 \over \pi } )+{cos((x+y)^2) \over \sqrt {2\pi}}) \end{align} $$
and exacly like first one you can write it for second one : $$ \begin{align} \int cos(y^2)C({\sqrt 2 (x+y) \over \sqrt \pi})dy=cos(y^2)((x+y)C((x+y) \sqrt { 2 \over \pi } )-{sin((x+y)^2) \over \sqrt {2\pi}}) \end{align} $$
and finally you just need to put all of this together and calculate them.