is $\int_1^x \sin(\sin(t)+\cos(t\sqrt{2}))\,dt$ a bounded function? The function $f(x)=\int_1^x \sin(\sin(t)+\cos(t\sqrt{2})) \, dt$ seems bounded, so I plot it in maple and I deduce that $f$ is always between $-1$ and $1$. I tried to prove it by searching the max and min of this function, but this doesn't help me at all. Do you any idea why $f$ is bounded?
 A: Let us notice a couple of things: the function $\varphi(t) = \sin(t)+\cos(t\sqrt{2})$ is not periodic but shares many properties with trigonometric polynomials: it is bounded, it is Lipschitz-continuous and it has mean zero, i.e. $\lim_{b\to +\infty}\frac{1}{b-a}\int_{a}^{b}\varphi(t)\,dt=0$. The same happens for any odd power of $\varphi(t)$. Since $\sin z$ is an odd entire function, $\sin z=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}z^{2n+1}$, we have:
$$ \int_{1}^{x}\sin\varphi(t)\,dt = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}\int_{1}^{x}\varphi(t)^{2n+1}\,dt. $$
It follows that, by denoting as $M_n$ the following supremum
$$ M_n = \sup_{x\geq 1}\left|\int_{1}^{x}\varphi(t)^{2n+1}\,dt\right| $$
to prove the existence of an absolute constant $D>0$ such that $M_n\leq D^n$ is enough to prove the boundedness of the original integral function. We may notice that the stationary points of $\int_{1}^{x}\varphi(t)^{2n+1}\,dt$ occur at the zeroes of $\varphi$, which are simple zeroes, pretty much evenly distributed along the real line. In particular the distance between a real zero of $\varphi$ and the next one never exceeds an absolute constant $E>0$. The absolute value of $\varphi$ is bounded by $2$ on the interval between a zero and the next one, hence $M_n\leq E\cdot 2^{2n+1}$ and
$$ \left|\int_{1}^{x}\sin\varphi(t)\,dt\right|\leq E\sinh(2),$$
which is a very crude inequality, but proves for sure the boundednes of the LHS.

In order to achieve better bounds, one may consider the Fourier series of $\widetilde{\varphi}^{2n+1}(t)$ where $\widetilde{\varphi}(t)=\sin(t)+\cos\left(\tfrac{p}{q}t\right)$ and $\frac{p}{q}$ is a convergent of the continued fraction of $\sqrt{2}=[1;2,2,2,\ldots]$, with the magnitude of $q$ being chosen according to the magnitude of $x$.
Things become quite more technical, but the main idea stays the same. 
