Prove $\int_{-\pi}^{\pi}\sin \sin x \,dx=0$ without using the fact that $\sin(x)$ is odd. Prove $$\large\int_{-\pi}^{\pi}\sin (\sin x) \,dx =0$$ without using the fact that $\sin(x)$ is odd.
Computing this in wolfram says that it is uncomputable, which leads me to believe that the only way to find this would be methods for solving definite integrals. I am wondering if it is possible with any other techniques such as DUIS or residues?

Another silly answer, using complex analytic methods (similar to sos440's answer, but avoiding use of Bessel functions):
Rewrite the integrand using Euler's formulas and put $z = e^{ix}$, thus mapping $[-\pi,\pi]$ to the unit circle (some algebra omitted):
$$
\int_{-\pi}^\pi \sin \sin x \, dx =
-\frac1{2} \int_{|z|=1}  \frac{\exp\Big( \frac12 ( z - \frac1z ) \Big) - \exp\Big( \frac12 ( -z + \frac1z ) \Big)}{z}\,dz.
$$
The integrand has an essential singularity at $z=0$, but we can still compute the relevant residue. Thanks to the $z$ in the denominator, we only have to compute the $0$:th terms of the Laurent series for the numerator.
We have
\begin{align}
\exp\Big( \frac12 ( z - \frac1z ) \Big) &=
e^{z/2} \cdot e^{-1/(2z)} \\
&= \Big( 1 + \frac{1}{1!} \big( \frac{z}{2} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 + \cdots \Big)
\Big( 1 - \frac{1}{1!} \big( \frac{1}{2z} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 - \cdots \Big)
\end{align}
Hence, the $0$:th term will be
$$
1 - \frac{1}{1!} \frac{1}{2^2} + \frac{1}{2!} \frac{1}{2^4} - \frac{1}{3!} \frac{1}{2^6} + \cdots
$$
Similarly
\begin{align}
\exp\Big( \frac12 (-z + \frac1z ) \Big) &=
e^{-z/2} \cdot e^{1/(2z)} \\
&= \Big( 1 - \frac{1}{1!} \big( \frac{z}{2} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 - \cdots \Big)
\Big( 1 + \frac{1}{1!} \big( \frac{1}{2z} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 + \cdots \Big)
\end{align}
And again, the $0$:th term will be
$$
1 - \frac{1}{1!} \frac{1}{2^2} + \frac{1}{2!} \frac{1}{2^4} - \frac{1}{3!} \frac{1}{2^6} + \cdots
$$
Summing up, the $0$:th term in the Laurent series for the numerator vanishes, and by the residue theorem, the integral, unsurprisingly is $0$.
Of course, this approach also uses, albeit implicitly, since I bothered to write out a number of unnecessary calculations, that the integrand is odd.
