Consider the following non-linear differential equation, $$ \dot{x}(t)=a-b\sin(x(t)), \ \ x(0)=x_0\in\mathbb{R}, $$ and assume that $a$ and $b$ are positive real numbers with $a>b$. Note that the solution $x(t)$ exists and can be analytically computed (see here).
My question: Is $\cos(x(t))$ a zero-mean periodic function of $t\ge 0$?
Further on, in case the answer to my previous question is in the affirmative, it should be that $$ \left|\int_{0}^t \cos(x(t))\, \mathrm{d}t\right|\le K(a,b), \ \ \forall t\ge 0, $$ where $K(a,b)$ is a positive constant depending on $a$ and $b$ only. So an additional question is: Can we find an explicit expression for $K(a,b)$?
N.B. Numerical simulations seem to confirm the above claims. However I couldn't quite prove them. So any help is really appreciated!
The change of variable $x(s)=u$ yields $$du=x'(s)\,ds=(a-b\sin u)\,ds$$ hence, for every $t$, $$\int_0^t\cos x(s)\,ds=\int_{x_0}^{x(t)}\frac{\cos u\,du}{a-b\sin u}=\frac1b\log\left(\frac{a-b\sin x_0}{a-b\sin x(t)}\right)$$ Thus the integral is indeed $0$ when $x(t)=x_0+2k\pi$ for some integer $k$, and it is extremal when $x_0=-\frac\pi2+k\pi$ and $x(t)=\frac\pi2+k\pi$ for some integer $k$ (maximal if $k$ is even and minimal if $k$ is odd), then its absolute value is
As a way of (very) partial confirmation of this formula, note that, by an elementary change of variable, $$K(a,b)=\frac1bK\left(\frac ab,1\right)$$