$\int_0^1 f(\sin(1/x)) \times g(\cos(1/x)) dx \leq \int_0^1 f(\sin(1/x)) dx \times \int_0^1 g(\cos(1/x))dx? $ I have noticed experimentally that the following question has a positive answer.
Is it true that for all even and convex functions $f$, $g$:
$$\int_0^1 f(\sin(1/x)) \times g(\cos(1/x)) dx \leq  \int_0^1 f(\sin(1/x)) dx \times \int_0^1 g(\cos(1/x))dx? $$
 A: $\newcommand\abs[1]{\lvert#1\rvert}$It already suffices that $f$ and $g$ be even and nondecreasing on $[0,1]$ (which of course is the case if $f$ and $g$ are even and convex). Indeed, then the identity $\abs\sin^2+\abs\cos^2=1$ implies that for all real $u$, $v$ we have
$$(\abs{\sin u}-\abs{\sin v})(\abs{\cos u}-\abs{\cos v})\le0\tag{1}\label{1}$$
and hence
$$\begin{aligned}h(u,v)&:=[f(\sin u)-f(\sin v)][g(\cos u)-g(\cos v)] \\ 
&=[f(\abs{\sin u})-f(\abs{\sin v})][g(\abs{\cos u})-g(\abs{\cos v})]\le0,
\end{aligned}$$
so that the difference between the left-hand side of your inequality and its right-hand side is
$$\frac12\,\int_0^1\int_0^1 dx\,dy\,h\Big(\frac1x,\frac1y\Big)\le0.
$$

One may note that the above reasoning holds if in the inequality in question one replaces all instances of $1/x$ by $k(x)$, where $k$ is any Borel-measurable function from $(0,1)$ to $\mathbb R$. Also, one can replace $\sin$ and $\cos$ by any functions $S$ and $C$ from $\mathbb R$ to $\mathbb R$ that are Borel-measurable and (say) bounded and "negatively dependent" in the sense that \eqref{1} holds with $S$ and $C$ in place of $\sin$ and $\cos$:
$$(\abs{S(u)}-\abs{S(v)})(\abs{C(u)}-\abs{C(v)})\le0
\tag{2}\label{2}$$
for all real $u$, $v$.
In particular, inequality \eqref{2} will hold if $|S|$ is any increasing function and $|C|$ is any decreasing one (or vice versa).
