Although an integral for $x=\dots$ exists, it is slightly harder to integrate. Dirac $\delta(t)$ helps solve $\cos(x)+ax=b$: $$\frac1{\sin(x)-a}=\int_a^b \delta(\cos(t)+at-b)dt\tag1$$ From numerical testing, $a,b$ have little effect on the integral convergence. Now use an $\delta(t)$ integral representation with $a=0,b=\pi,\frac\pi2$. The integration order is interchangeable and comparing a double integral with its single delta integral shows that $(2)$ probably approaches $(1)$
$$\frac1{\sin(x)-a} =\frac1\pi\int_0^\pi\int_0^\infty\cos(w (\cos(t)+a t-b))dwdt= \frac1\pi\int_0^\infty\int_0^\pi\cos(w (\cos(t)+a t-b))dtdw\tag2$$ Now we see a familiar Bessel,Struve H$_v(x)$ or Anger-Weber integral representation, like for the Kepler equation series, which is a slightly different integrand from the problem we have. Other partial results are: $$\int_0^\frac\pi2 \cos(u\cos(t)+t)dt=\frac\pi2\text H_{-1}(u)+\frac{\cos(u)-1}u$$ if one uses a $\cos(a+b)$ formula, then we have the hypergeometric $_1\text F_2(a;b,c;z)$, likely in terms of the above functions: $$\int_0^\frac\pi2 \cos(at)\cos(b\cos(t))dt=\frac1a\sin\left(\frac{ab}2\right)\,_1\text F_2\left(1;1-\frac a2,1+\frac a2;-\frac{b^2}4\right)$$ Integration by parts on the integrand does not remove the $t$ making application of similar formulas to the above fail. What is a closed form of $\displaystyle\int_0^{\pi\text{ or }\frac\pi2}\cos(w (\cos(t)+a t-b))dt $?
