Integral of Infinite Sines I constructed the following question:
Let $S_n$ denote the sequence where
\begin{align}
S_1=&\sin{x}\\
S_2=&\sin{(\sin{x})}\\
S_3=&\sin(\sin(\sin{x}))\\
&\vdots
\end{align}
Evaluate
$$I=\int_0^{\pi}\lim_{n\rightarrow\infty}S_n\,dx$$
Messing around in desmos, it would seem that $S_n$ approaches $0$ as $n\rightarrow\infty$. However I cant seem to be able to prove this. Any ideas?
 A: For $x_1 \in [0,\pi]$ the sequence defined by $ x_{n+1} = \sin x_n$ and $x_1$ converges to $0$: it is non negative and non increasing, hence converges. As $x \mapsto \sin x$ is continuous and $0$ is its only fix point, the limit is equal to zero.
Hence $ S_n$ converges pointwise to zero on $[0,\pi]$ and $$I=\int_0^{\pi}\lim_{n\rightarrow\infty}S_n\,dx = 0$$
A: The limit is where $S_n$ stops changing, in other words where the previous evaluation equals the next:
$$S_{n+1} "=" S_n \implies x = \sin x$$
which only has the solution $x=0$
Let's analyze this slightly more rigorously. Denote the sequential "derivative" of $S_n$ by $$S_{n+1} - S_n = \sin(S_n) - S_n$$
Since this gives the relationship between consecutive terms, we can figure out where the sequence is increasing or decreasing. For $S_n > 0$, the sequence will always decrease next, and for $S_n > 0$, it will always increase next.
Now we don't necessarily have monotonicity for arbitrary starting points, but consider that 
$$|S_{n+1}| = |\sin(S_n)| < |S_n|$$
hence it will always grow closer to $0$ for any initial starting value in $\mathbb{R}$. Since we already proved that $0$ was a fixed point, we are done.
