Periodicity of the solution of a non-linear differential equation 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!
 A: As in my response to your question Behavior of a non-linear differential equation on MO, consider the ODE on the one-dimensional torus (that is, circle) $\mathbb{T} = \mathbb{R}/2 \pi \mathbb{Z}$.  By (ergodic theoretical version of) Liouville's theorem (see, e.g., Theorem 1 on p. 48 of Ergodic Theory by Cornfeld, Fomin and Sinai) the probability measure
$$
d\mu = \frac{\sqrt{a^2 - b^2}}{2 \pi} \frac{1}{a - b \sin{x}}\, d\lambda,
$$
where $\lambda$ is one-dimensional Lebesgue measure on $\mathbb{T}$, is an invariant (probability) measure for the flow $\Phi$ generated by the ODE on $\mathbb{T}$.  Further, the flow $\Phi$, being a reparametrization of the flow generated on $\mathbb{T}$ by $\dot{x} \equiv 1$, is uniquely ergodic.  
By the (continuous time version of) Birkhoff ergodic theorem for uniquely ergodic DSs (see, e.g., Uniform convergence of Birkhoff averages and unique ergodicity on MO), for every continuous function $f \colon \mathbb{T} \to \mathbb{C}$ its time averages
$$
\frac{1}{t} \int\limits_{0}^{t} f(\Phi(s, x_0))\, ds
$$
converge, as $t \to \infty$, uniformly in $x_0 \in  \mathbb{T}$, to its space average
$$
\int\limits_{\mathbb{T}} f(x) \, d\mu(x).
$$
In our case we have
$$
\lim\limits_{t\to\infty} \frac{1}{t} \int\limits_{0}^{t} f(\Phi(s, x_0))\, ds = \frac{1}{T} \int\limits_{0}^{T} \Phi(s,x_0) \, ds,
$$
where $T = 2\pi/\sqrt{a^2 - b^2}$ is the period.  
Taking $f(x) = \cos{x}$ we have
$$
\int\limits_{\mathbb{T}} f(x) \, d\mu(x) = \frac{\sqrt{a^2 - b^2}}{2 \pi} \int\limits_{0}^{2\pi} \frac{\cos{x}}{a - b \sin{x}} dx = 0.
$$
This is a (hopefully correct) answer to your first question.  Regarding the additional question, one can take just one solution (for instance, corresponding to $x_0 = 0$), so there is $K = K(a, b)$, but I do not know how easily one can find an explicit expression.
A: 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

$$K(a,b)=\frac1b\log\left(\frac{a+b}{a-b}\right)$$

As a way of (very) partial confirmation of this formula, note that, by an elementary change of variable, $$K(a,b)=\frac1bK\left(\frac ab,1\right)$$
