<s>[INST] An Inequality concerning double integral 
Prove $$ \frac{\pi}{8}\left(1-\cos\frac{2}{\pi}\right)\le \iint_D \sin (x^2)\cos (y^2) dxdy \le \frac{\pi}{8}(1-\cos 1),$$where the integrating region $D$ is enclosed by $x=0, y=0$ and $x+y=1$.

We can obtain
$$\iint_D \sin (x^2)\cos (y^2) dxdy=\int_0^1\int_0^{1-y}\sin(x^2)\cos(y^2)dxdy\le \int_0^1\int_0^{1-y}\sin(x^2)dxdy,$$
but which is still difficult.
 [/INST] We want to bound
$$
I = \iint_D \sin (x^2)\cos (y^2) dxdy 
$$
where the integrating region $D$ is enclosed by $x=0, y=0$ and $x+y=1$. Since the integrand is symmetric in $x$ and $y$, we can write this as
$$
I = \frac12 \iint_E \sin (x^2)\cos (y^2) dxdy 
$$
where the integrating region $E$ is enclosed by the rectangle  $x+y=0$,  $x+y=1$, $x-y=-1$,  $x-y=1$. Changing variables to $x+y=a$, $x-y=b$ gives
$$
I = \frac14 \int_{a=0}^1 \int_{b=-1}^1 \sin ((\frac{a+b}{2})^2)\cos ((\frac{a-b}{2})^2) dadb 
$$
Using the trig theorem $\sin(A) \cos(B) = 1/2 (\sin(A - B) + \sin(A + B))$ gives
$$
I = \frac18 \int_{a=0}^1 \int_{b=-1}^1 (\sin(\frac{a^2+b^2}{2}) - \sin(a b)) dadb 
$$
The integration of $\sin(a b)$ gives zero, and the trig theorem $\sin(A+B) = \cos(B) \sin(A) + \cos(A) \sin(B)$ gives us
$$
I = \frac12 \int_{a=0}^1 \sin(\frac{a^2}{2}) da \int_{b=0}^1\cos(\frac{b^2}{2})  db 
$$
Note that the integrals now factorize, and that so far we have made no approximations at all.
We can now compute bounds. Since in the intervals of integration, $ \sin(\frac{a^2}{2})$ is convex and  $\cos(\frac{b^2}{2})$ is concave,  we have
$$
I \le \frac12 \cdot \frac{\sin(1/2)}{2} \cdot 1 = \frac{\sin(1/2)}{4} \simeq 0.12 < \frac{\pi}{8}(1-\cos 1) \simeq 0.18 
$$
For the lower bound, we use the tangent slope of $f(a) = \sin(\frac{a^2}{2})$ , which is $f'(a) = a \cos(\frac{a^2}{2})$, at $a=1$, to get
$$
I \ge \frac12 \cdot \frac{(\sin(1/2))^2}{2 \cos(1/2)} \cdot \frac{1 + \cos(1/2)}{2}  \simeq 0.0615 
$$
Now this is reasonable bound, however it doesn't prove the conjecture since
$
 0.0615 \le \frac{\pi}{8}\left(1-\cos\frac{2}{\pi}\right) \simeq 0.077
$ . It is unlikely anyway that one should be able to prove the conjectured lower bound by reasonable approximations, since numerically, $I \simeq 0.079$, hence the lower bound is very tight, it is about $0.964$ times the true value. For comparison, the conjectured upper bound is not tight at all.
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