The goat problem has the following transcendental equation:
$$\sin(a)-a\cos(a)=\frac\pi2,a=1.905695729\dots$$
If $f^{-1}(x)$ is odd, its Fourier sine series of period $\left[-\frac L2,\frac L2\right]$ is: $$f^{-1}(x)=\frac2L\sum_{n=1}^\infty\sin(nx)\int_0^L f^{-1}(t)\sin\left(\frac{\pi nt}L\right)dt\mathop=^{f^{-1}(t)\to t}\frac2L\sum_{n=1}^\infty\sin(nx)\int_{f^{-1 }(0)}^{f^{-1}(L)}t\sin\left(\frac\pi Lf(t)n\right)df(t)$$
We use it to invert $\sin(x)-x\cos(x)$:
Therefore $x=\frac\pi2$ and:
$$\boxed{a=\frac2\pi\sum_{n=1}^\infty\sin\left(\frac{\pi n}2\right)\int_0^\pi t^2\sin(t)\sin(n(\sin(t)-t\cos(t)))dt}$$
We likely do not switch the sum and integral, but we use the sine difference formula for the integral:
$$\int_0^\pi t^2\sin(t)\sin(n(\sin(t)-t\cos(t)))dt=\int_0^\pi t^2\sin(t)(\cos(nt\cos(t))\sin(n\sin(t))-\cos(n\sin(t))\sin(nt\cos(t))dt$$
Which is reminiscent of Bessel, or Glasser for fun, function integral representations. Now we may expand $\sin(y)$ as a series if really needed. Is there any non-integral form of $\displaystyle\int_0^\pi t^2\sin(t)\sin(n(\sin(t)-t\cos(t)))dt$?
