Evaluating integrals of infinitely concatenated functions viz. Dominated Convergence Recently I devised an integral to solve: Let $a_{n}(x)=\arcsin\Big[\cos\big(x\cdot a_{n-1}(x)\big)\Big]$ with $a_{1}(x)=\arcsin\big[\cos(x)\big]$, evaluate 
$$\lim\limits_{n\to\infty}\int_{0}^{\frac{\pi}{4}}a_{n}(x)\mathrm dx$$
Here's my solution. Let $f=\lim\limits_{n\to\infty}x\cdot a_{n}(x)$, then 
$$\frac{1}{f}\arcsin\big[\cos(f)\big]=\frac{1}{x}\longrightarrow x=\frac{f}{\arcsin\big[\cos(f)\big]}$$
Without needing to explicitly compute it, define a function $\mathcal{A}(x)=f$ as the inverse to the function above. Then, 
$$\lim\limits_{n\to\infty}\int_{0}^{\frac{\pi}{4}}a_{n}(x)\mathrm dx=\int_{0}^{\frac{\pi}{4}}\frac{\mathcal{A}(x)}{x}\mathrm dx$$
In order to remove the inverse function, we make the substitution $x\to\frac{\theta}{\arcsin\big[\cos(\theta)\big]}$...and, for simplification purposes, let $p(\pi)=\frac{\pi^{2}}{2(\pi +4)}$, hence
$$\int_{0}^{\frac{\pi}{4}}\frac{\mathcal{A}(x)}{x}\mathrm dx=\int_{0}^{p(\pi)}\frac{\theta}{\frac{\theta}{\arcsin\big[\cos(\theta)\big]}}\cdot\Bigg[\frac{1}{\arcsin\big[\cos(\theta)\big]}+\frac{\theta\DeclareMathOperator{\sgn}{sgn}\sgn\big[\sin(\theta)\big]}{\arcsin^{2}\big[\cos(\theta)\big]}\Bigg]\mathrm d\theta$$
$$=p(\pi)+\int_{0}^{p(\pi)}\frac{\theta}{\arcsin\big[\cos(\theta)\big]}\mathrm d\theta$$
$$=p(\pi)+\int_{0}^{p(\pi)}\frac{\theta}{\frac{\pi}{2}- \theta}\mathrm d\theta$$
$$=p(\pi) + \dfrac{{\pi}\ln\left({\pi}\right)}{2}-\dfrac{\left({\pi}^2+4{\pi}\right)\ln\left(\frac{4{\pi}}{{\pi}+4}\right)+{\pi}^2}{2{\pi}+8}\approx 0.910499$$
This is a pretty convoluted solution, and I've learned that there's another way to simplify the integral through Dominated Convergence...would really like to see how that works!
 A: Your sequence may be recast as
$$ a_{n+1}(x) = \arcsin(\cos(x a_n(x))), \qquad a_0(x) = 1. $$
Now by the mathematical induction, we can check that $ 0 \leq a_n(x) \leq \frac{\pi}{2} $ holds for all $n \geq 0$ and $0 \leq x \leq \frac{\pi}{4}$. Indeed, this is obviously true for $n = 0$, and assuming that this is true for $n \geq 0$, then
\begin{align*}
&0 \leq a_n(x) \leq \tfrac{\pi}{2}\quad\text{and}\quad 0\leq x\leq\tfrac{\pi}{4} \\
&\Rightarrow\qquad 0 \leq x a_n(x) \leq \tfrac{\pi}{2} \\
&\Rightarrow\qquad 0 \leq \cos(x a_n(x)) \leq 1 \\
&\Rightarrow\qquad 0 \leq a_{n+1}(x) \leq \tfrac{\pi}{2}.
\end{align*}
Now using the identity $\arcsin(\cos(x)) = \frac{\pi}{2}-x$ which holds true for $0\leq x\leq\frac{\pi}{2}$, we get
$$ a_{n+1}(x) = \tfrac{\pi}{2} - x a_n(x). $$
Using this, we easily check that
$$ \left| a_{n+2}(x) - a_{n+1}(x) \right| \leq \tfrac{\pi}{4} \left|a_{n+1}(x) - a_{n}(x)\right|, $$
which then shows that $a_n(x)$ converges uniformly over $[0, \frac{\pi}{4}]$. Moreover, if $a(x) := \lim_{n\to\infty} a_n(x)$ denotes the limit function, then it must satisfy $a(x) = \tfrac{\pi}{2} - x a(x)$, from which we deduce
$$a(x) = \frac{\pi}{2(x+1)}.$$
Therefore by a suitable convergence theorem (such as the convergence theorem for Riemann integral under uniform convergence or the dominated convergence theorem), we get
$$ \lim_{n\to\infty} \int_{0}^{\frac{\pi}{4}} a_n(x) \, \mathrm{d}x
= \int_{0}^{\frac{\pi}{4}} a(x) \, \mathrm{d}x
= \frac{\pi}{2}\log\left(1+\frac{\pi}{4}\right) \approx 0.9104986622 \cdots.$$
