<s>[INST] 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?
 [/INST] 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.
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