To formally justify this integration, we can do the following. Put $f_1(x) = \sqrt x$ ,and for each $n\ge 1$, put $f_{n+1}(x) = \sqrt{x + f_{n}(x)}$. We want to show that $f(x) = \lim_{n\to\infty}f_n(x)$ exists and is integrable on $[0,2]$. We will show that $f(x)$ is continuous on $(0,2]$, and has a discontinuity at $x = 0$, where $f(0) = 0$. This is sufficient because a function $f\colon[0,2]\to\mathbf R$ is Riemann integrable if and only if it is bounded and continuous almost everywhere. Finally, by "correcting" the value of $f$ at $x=0$ to produce a continuous function $\tilde f$ on $[0,2]$, we can evaluate the integral $\int_0^2f(x)\,\mathrm dx$ by applying the fundamental theorem of calculus to integrate $\int_0^2\tilde f(x)\,\mathrm dx$.
For each $x\in (0,2]$, put $x_n = f_n(x)$. We claim that $x_n\to \frac{1}{2} + \frac{1}{2}\sqrt{4x+1}$. First we will justify the convergence of the sequence $(x_n)$ using the monotone convergence theorem and then use algebraic limit laws to deduce its limit rigorously.
Consider the sequence $b_{n+1} = \sqrt{2 + b_n}$, where $b_1 = \sqrt 2$. Clearly $b_1 \leqslant 2$. Suppose that $b_n\leqslant 2$; then $\sqrt{2 + b_n} \leqslant \sqrt{2 + 2} = 2$. By induction, $b_n \leqslant 2$ for every $n$. Thus the sequence $(b_n)$ is bounded above by $2$. The sequence $(b_n)$ is clearly monotone increasing. Hence $(b_n)$ converges.
Since $(x_n)$ is monotone and bounded above by $\lim_{n\to\infty}b_n$, it converges, so set $L = \lim_{n\to\infty}x_n$. Since $(x_n)$ converges to $L$, every subsequence converges to $L$, so by continuity of the map $y\mapsto y^2$, we have
$$
L^2 = \big(\lim_{n\to\infty}x_{n+1}\big)^2 = \lim_{n\to\infty}x_{n+1}^2 = \lim_{n\to\infty}x + x_n = x + L.
$$
Therefore $L$ is a root of the polynomial $y\mapsto y^2 - y - x$, and we can conclude that
$$
L = \frac{1}{2} + \frac{1}{2}\sqrt{4x+1}.
$$
(Note that $L$ could not be the negative root since $x_n \geqslant 0$ for every $n\geqslant 1$ and every $x\in(0,2]$.)
If $x = 0$, then $x_n = 0$ for each $n$, so all told,
$$
f(x) = \begin{cases}
\frac{1}{2} + \frac{1}{2}\sqrt{4x+1}, &\text{if $x\in (0,2]$,}\\
0, &\text{if $x = 0$.}
\end{cases}
$$
Thus we have proved that the integrand $f(x)=\sqrt{x+\sqrt{x+\dotsb}}$ is bounded and continuous almost everywhere on $[0,2]$, so it is Riemann integrable on $[0,2]$. The function $\tilde f\colon [0,2]\to\mathbf R$ defined by
$$
\tilde f(x) = \frac{1}{2} + \frac{1}{2}\sqrt{4x+1},\quad\text{for each $x\in[0,2]$,}
$$
is continuous on $[0,2]$ and agrees with $f$ almost everywhere, so $\int_0^2 f(x)\,\mathrm dx = \int_0^2\tilde f(x)\,\mathrm dx$. Since $\tilde f(x)$ is continuous on the entire closed interval $[0,2]$, its integral can be evaluated via the fundamental theorem of calculus. Since
$$
F(x) = \frac{x}{2} + \frac{(4x+1)^{3/2}}{12}
$$
satisfies $F'(x) = \tilde f(x)$ for each $x\in [0,2]$, by the fundamental theorem of calculus,
\begin{align*}
\int_0^2\tilde f(x)\,\mathrm dx = F(2)-F(0) = \bigg[\frac{2}{2} + \frac{(4\cdot 2+1)^{3/2}}{12}\bigg] - \bigg[\frac{0}{2}+\frac{(4\cdot0+1)^{3/2}}{12}\bigg]=\frac{19}{6},
\end{align*}
as desired.
In this answer it is proved that if $f, g$ are Riemann integrable on $[0,T]$ and $g$ is periodic with period $T$ then $$\lim_{n\to\infty} \int_{0}^{T}f(x)g(nx)\,dx=\frac{1}{T}\left(\int_{0}^{T}f(x)\,dx\right)\left(\int_{0}^{T}g(x)\,dx\right)$$ For your problem we have $$f(x) =\sin x, g(x) =\frac{1}{1+\cos^2x}$$ and hence the desired limit is $$\frac{1}{\pi}\int_{0}^{\pi}\sin x\, dx\int_{0}^{\pi}\frac{dx}{1+\cos^2x}=\frac{4}{\pi}\int_{0}^{\pi}\frac{dx}{3+\cos x}=\frac{4}{\pi}\cdot\frac{\pi}{2\sqrt{2}}=\sqrt{2}$$
Best Answer
Consider the limit piecewise
$$\lim_{n\to\infty}x^{2n}=\begin{cases}0 & |x| < 1 \\ +\infty & |x| > 1\end{cases}$$
which by the continuity of $e^x$ gives the following limits
$$\lim_{n\to\infty}\exp(-x^{2n}) = \begin{cases}e^0 = 1 & |x| < 1 \\ e^{-\infty} = 0 & |x| > 1\end{cases}$$
so the integral converges to
$$\lim_{n\to\infty}\int_{-\infty}^\infty e^{-x^{2n}}\:dx = \int_{-1}^1 dx = 2$$
by dominated convergence.