Prove or refute: If $f$ is Riemann integrable on $[0,1]$, then so is $\sin(f)$ I'd love your help with the following question: I need to prove or refute the claim that for a Riemann integrable function $f$ in $[0,1]$ also $\sin(f)$ is integrable on $[0,1]$.
My translation for this claim: If $\int_{0}^{1} f(x) dx < \infty$, so does $\int_{0}^{1} \sin(f(x))dx < \infty$, Am I right? 
I tried to think of an elementary function that will fit the conditions, one that will blow up in $0$ or $1$ or both, but I didn't find any. Can I just use the fact that $\int_{0}^{1} \sin(f(x))dx \leq \int_{0}^{1} 1dx < \infty$ and that's it or Am I missing something?
Thanks!

I do not think your translation is correct (unless you meant Lebesgue, and not Riemann integrable). The concept of Riemann integrable and Lebesgue integrable are not the same.
Riemann integrable: $f\colon[a,b]\to\mathbb{R}$, $-\infty<a<b<+\infty$, $f$ bounded and the upper integral equal to the lower integral. 
Lebesgue integrable: $f\colon E\subset\mathbb{R}\to\mathbb{R}$, $E$ Lebesgue measurable set, $f$ Lebesgue measurable and $\int_E|f|<\infty$.
There are several ways of showing that if $f\colon[0,1]\to\mathbb{R}$ is Riemann integrable so is $\sin(f)$. The easiest way is to use the following fact: a bounded function $g\colon[a,b]\to\mathbb{R}$ is Riemann integrable if and only if the set of points where $g$ is discontinuous has measure $0$.
First of all it is cleat that $\sin(f)$ is bounded. Since $\sin$ is a continuous function,
$$
\{x\in[0,1]:\sin(f)\text{ is discontinuous at }x\}\subset\{x\in[0,1]:f\text{ is discontinuous at }x\}.
$$
Since $f$ is Riemann integrable, the set on the right hand side is of measure $0$, and so are all its subsets.
