We give the following auxiliary results.
Fact 1: $\sin u \le u - \frac{1}{6}u^3 + \frac{1}{120}u^5$ for all $u \ge 0$.
Fact 2: $x^{-x} \le \frac{3 - x}{x^2 - x + 2} \le 1$ for all $x \in [1, 2]$.
Fact 3: $x^{-x} \le a^{-x} \mathrm{e}^{-x + a}$ for all $x, a > 0$.
(Proof: It is equivalent to $\ln u \ge 1 - \frac{1}{u}$ for all $u > 0$.)
Using Fact 1, we have (cf. Sophomore's dream)
\begin{align*}
&\int_0^1 \sin (x^{-x})\,\mathrm{d}x \\
\le\,& \int_0^1 \left(x^{-x} - \frac16 x^{-3x} + \frac{1}{120}x^{-5x}\right)\mathrm{d}x\\
=\,& \sum_{n=0}^\infty \frac{1}{(n + 1)^{n + 1}} - \frac16 \sum_{n=0}^\infty \frac{3^n}{(n + 1)^{n + 1}} + \frac{1}{120}
\sum_{n=0}^\infty \frac{5^n}{(n + 1)^{n + 1}}\\
=\, & \sum_{n=0}^5 \frac{1 - 6^{-1}3^n + 120^{-1}5^n}{(n + 1)^{n + 1}}
+ \sum_{n=6}^\infty \frac{1 - 6^{-1}3^n + 120^{-1}5^n}{(n + 1)^{n + 1}}\\
<\, & \sum_{n=0}^5 \frac{1 - 6^{-1}3^n + 120^{-1}5^n}{(n + 1)^{n + 1}}
+ \sum_{n=6}^\infty \frac{1 - 6^{-1}3^n + 120^{-1}5^n}{7^{n + 1}}\\
=\,& \frac{62861674901693}{65868380928000}. \tag{1}
\end{align*}
Using Facts 1-2, we have
\begin{align*}
&\int_1^2 \sin(x^{-x})\, \mathrm{d} x \\
\le\,& \int_1^2 \sin\left(\frac{3 - x}{x^2 - x + 2}\right)\,\mathrm{d} x\\
\le\,& \int_1^2 \left[\frac{3 - x}{x^2 - x + 2} - \frac{1}{6}\left(\frac{3 - x}{x^2 - x + 2}\right)^3 + \frac{1}{120}\left(\frac{3 - x}{x^2 - x + 2}\right)^5\right]\mathrm{d}x\\
=\,& \frac{625\sqrt7}{1029}
\arctan\frac{\sqrt7}{5} + \frac{1687723}{18063360} - \frac12\ln 2. \tag{2}
\end{align*}
Using Fact 3, we have
$$\int_4^5 \sin (x^{-x})\, \mathrm{d}x
\le \int_4^5 x^{-x}\, \mathrm{d} x
\le \int_4^5 4^{-x} \mathrm{e}^{-x + 4}\, \mathrm{d} x = \frac{4 - \mathrm{e}^{-1}}{1024 + 2048\ln 2}, \tag{3}$$
and
$$\int_3^4 \sin (x^{-x})\, \mathrm{d}x
\le \int_3^4 x^{-x}\, \mathrm{d} x
\le \int_3^4 3^{-x} \mathrm{e}^{-x + 3}\, \mathrm{d} x = \frac{3 - \mathrm{e}^{-1}}{81 + 81\ln 3}, \tag{4}$$
and
$$\int_{5/2}^3 \sin(x^{-x})\,\mathrm{d} x
\le \int_{5/2}^3 x^{-x}\,\mathrm{d} x
\le \int_{5/2}^3 (5/2)^{-x}\mathrm{e}^{-x + 5/2}\,\mathrm{d} x = \frac{4\sqrt{10} - 8\sqrt{\mathrm{e}^{-1}}}{125\ln \frac{5}{2} + 125}, \tag{5}$$
and
$$\int_2^{5/2} \sin(x^{-x})\,\mathrm{d} x
\le \int_2^{5/2} x^{-x}\,\mathrm{d} x
\le \int_2^{5/2} 2^{-x}\mathrm{e}^{-x + 2}\,\mathrm{d} x = \frac{2 - \sqrt{2\mathrm{e}^{-1}}}{8 + 8\ln 2}. \tag{6}$$
Also, we have
$$\int_5^\infty \sin (x^{-x})\, \mathrm{d}x
\le \int_5^\infty x^{-x}\, \mathrm{d}x
\le \int_5^\infty \mathrm{e}^{-x\ln 5}\, \mathrm{d}x = \frac{1}{3125\ln 5}. \tag{7}$$
With the above results, we obtain the desired result
$$\int_0^\infty \sin(x^{-x})\,\mathrm{d}x < \frac{1 + \sqrt5}{2}.$$
Note: $(1) + (2) + (3) + (4) + (5) + (6) + (7)$ gives $\int_0^\infty \sin(x^{-x})\,\mathrm{d}x < 1.617374660$.
We are done.
We need to prove that
$$\int_0^1 \frac{1}{x^2 + \ln^2 x}\, \mathrm{d} x > 1.$$
Fact 1: For all $y \in (0, \mathrm{e}]$,
$$\frac{-y^4 + 64y^3 - 318y^2 + 400y + 35}{132y + 48} \ge y(1 - \ln y)^2.$$
(Note: LHS is the (4, 1)-Pade approximant of $y(1 - \ln y)^2$ at $y = 1$.)
With the substitution $y = \mathrm{e}x$, using Fact 1, we have
\begin{align*}
& \int_0^1 \frac{1}{x^2 + \ln^2 x}\, \mathrm{d} x\\
=\, & \int_0^\mathrm{e} \frac{\mathrm{e}}{y^2 + \mathrm{e}^2 (1 - \ln y)^2}\, \mathrm{d} y\\
\ge\, & \int_0^\mathrm{e} \frac{132\mathrm{e}y^2 + 48\mathrm{e}y}{(132 - \mathrm{e}^2)y^4 + (64\mathrm{e}^2 + 48)y^3 - 318\mathrm{e}^2 y^2 + 400\mathrm{e}^2y + 35\mathrm{e}^2}\, \mathrm{d} y\\
\ge\, & \int_0^\mathrm{e} \frac{132\mathrm{e}y^2 + 48\mathrm{e}y}{(132 - \mathrm{e}^2)(y + 1/12)(y + 463/64)[(y - 69/44)^2 + (55/54)^2]}\, \mathrm{d} y \tag{1} \\[8pt]
=\, &
\frac{1253273472\mathrm{e}}{911936989\mathrm{e}^2 - 120375682548} \int_0^\mathrm{e} \frac{1}{12y + 1}\, \mathrm{d} y\\[8pt]
&\quad + \frac{29129459463143424\mathrm{e}}{38952637956653\mathrm{e}^2 - 5141748210278196}
\int_0^\mathrm{e} \frac{1}{64y + 463}\, \mathrm{d} y \\[8pt]
&\quad + \frac{372594816\mathrm{e}}{2487336612548656836 - 18843459185974673\mathrm{e}^2}\\[8pt]
&\quad\quad \times \int_0^\mathrm{e} \frac{842186377401732y + 746090344086823}{1411344y^2 - 4426488y + 4934869}\, \mathrm{d} y \tag{2}\\
>\, & 1
\end{align*}
where in (1) we have used (easy): for all $y\in [0, \mathrm{e}]$,
\begin{align*}
&(132 - \mathrm{e}^2)y^4 + (64\mathrm{e}^2 + 48)y^3 - 318\mathrm{e}^2 y^2 + 400\mathrm{e}^2y + 35\mathrm{e}^2\\
\le\, & (132 - \mathrm{e}^2)(y + 1/12)(y + 463/64)[(y - 69/44)^2 + (55/54)^2].
\end{align*}
Note: The three integrals in (2) admit closed form expressions in terms of $\ln $ and $\arctan$ only.
We are done.
Best Answer
Let's approximate $\sin(-20)$ using a Taylor series centered near $-20$. Since $\pi \approx 3.14$, we have $-6.5\pi \approx -20.41$, so this is what we'll use as the center of the approximation. Note that $\sin(-6.5\pi) = -1$ and $\cos(-6.5\pi) = 0$. So we have $$\sin(x) = \sum_{k=0}^{\infty} (-1)^k \frac{\sin(-6.5\pi)}{(2k)!} \cdot (x + 6.5\pi)^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(2k)!} (x + 6.5\pi)^{2k}.$$ We'll use the quadratic approximation: $$\sin(x) = -1 + \frac{1}{2} \cdot (x + 6.5\pi)^2 + O((x + 6.5\pi)^4).$$ Since $3.14159 < \pi < 3.14160$, we have $0.42 < -20 + 6.5\pi < 0.421$. We compute $$\sin(-6.5\pi + 0.42) \approx -1 + \frac{1}{2} \cdot 0.42^2 = -0.9118$$ and $$\sin(-6.5\pi + 0.421) \approx -1 + \frac{1}{2} \cdot 0.421^2 = -0.9113795.$$ By Taylor's theorem, these approximations have error bounded above by $0.421^4/4! < 0.002$. So $$-0.914 < \sin(-6.5\pi + 0.42) < \sin(-20) < \sin(-6.5\pi + 0.421) < -0.909.$$ Thus, $$20.086 < \sin(-20) + 21 < 20.091.$$
At this point, we could try to approximate the natural logarithm, but its Taylor series converges rather slowly, so it's more efficient to bound $e^3$. We have $e < 2.7183$, so $$e^3 < 2.7183^3 = 20.08593... < 20.086 < \sin(-20) + 21.$$ Thus $3 < \ln(\sin(-20) + 21)$.