Find the sign of $\int_{0}^{2 \pi}\frac{\sin x}{x} dx$

$$ \begin{align*} \int_0^{2\pi}\frac{\sin x}{x}\,dx&=\int_0^{\pi}\frac{\sin x}{x}\,dx+\int_\pi^{2\pi}\frac{\sin x}{x}\,dx\\ &=\int_0^{\pi}\frac{\sin x}{x}\,dx+\int_0^{\pi}\frac{\sin(x+\pi)}{x+\pi}\,dx\\ &=\int_0^{\pi}\Bigl(\frac{1}{x}-\frac{1}{x+\pi}\Bigr)\sin x\,dx\\ &=\pi\int_0^{\pi}\frac{\sin x}{x(x+\pi)}\,dx\\ &>0 \end{align*} $$

Regarding the sign, it is easy the check that every area in each $\pi$ interval is always smaller than the preceeding one. The sign is positive.

For the value, integrate in the same interval

$$y = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$$

The difference between the sinc function and that is at most $\approx 0.015$ in that interval.

Adding $$\cos \frac{x}{16}$$ makes the error at most $\approx 0.003$

For the first one you have.

$$I = \frac{104}{105} \sqrt{2} \sim\sqrt{2} $$

For the latter:

$$I = \frac{{1568}}{{2145}}\frac{{\sqrt {2 + \sqrt 2 } }}{2} \sim \sqrt{2}$$

Maybe the area is $\sqrt{2}$ after all.