Trouble with an Odd Integral Here is a Integral I am having some trouble with
$$\int_0^\pi \sin^2\Big(x^2-\sqrt{\pi^2-x^2}\Big)dx$$
A while back I integrated $\int_0^\pi \sin^2\Big(x-\sqrt{\pi^2-x^2}\Big)dx$ and found it was equal to $\frac{\pi}{2}$; so I tried  to substitute $u=x^2$  that got me somewhere, but I am having quite a bit of trouble. Any help would be much appreciate
 A: Let us first prove this statement:
$$\int_0^\pi \sin^2\Big(x-\sqrt{\pi^2-x^2}\Big)dx=\frac{\pi}{2}$$
Actually, we can compute a more general integral:
$$I(2n)=\int_0^\pi \sin^{2n}\Big(x-\sqrt{\pi^2-x^2}\Big)dx$$
$$n=0,1,2,...$$
Let's split this integral into two:
$$I_1(2n)=\int_0^\frac{\pi}{\sqrt{2}} \sin^{2n}\Big(\sqrt{\pi^2-x^2}-x\Big)dx$$
$$I_2(2n)=\int_\frac{\pi}{\sqrt{2}}^\pi \sin^{2n}\Big(x-\sqrt{\pi^2-x^2}\Big)dx$$
$$I(2n)=I_1(2n)+I_2(2n)$$
A bound of integration $\frac{\pi}{\sqrt{2}}$ is a positive solution of the equation:
$$x-\sqrt{\pi^2-x^2}=0$$
Now, make a change of variable in the integrals:
$$\sqrt{\pi^2-x^2}-x=t$$ for $I_1(2n)$
and
$$x-\sqrt{\pi^2-x^2}=t$$ for $I_2(2n)$
After the elementary transformations, we reach the classical integral:
$$I(2n)=\int_0^\pi \sin^{2n}t \;dt=\frac{(2n)!}{(2^nn!)^2}\pi$$
Now, what about
$$I=\int_0^\pi \sin^2\Big(\sqrt{\pi^2-x^2}-x^2\Big)dx$$
The same method as the previous one is not suitable because we need to calculate integrals like
$$\int \sin x^2\;dx$$
$$\int \cos x^2\;dx$$
Below is a physicist's approach to get an approximate analytical solution.
The only way to achieve a such result is to linearize the argument of the sine function.
I use the simplest linear approximation
$$\sqrt{\pi^2-x^2}-x^2\approx \pi-(1+\pi)x$$
$$0\leqslant x \leqslant\pi$$
Result
$$I_{approx}=\frac{\pi}{2}-\frac{\sin(2\pi^2)}{4(1+\pi)}=1.52...$$
Exact value of the integral is $1.36...$ The approximation error is about $0.16$
For a physicist this is quite a satisfactory result.
