$(1)\hspace{5mm} \displaystyle 1.5<\int\limits_0^\infty\frac{dx}{1+x^2 |\sin x|} =\sum\limits_{k=0}^\infty \int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x} $$\displaystyle < 1.52 + \sum\limits_{k=1}^\infty \int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x}$
because of $\enspace \displaystyle 1.5<\int\limits_0^\pi\frac{dx}{1+(-1)^k x^2 \sin x}<1.52 \enspace$ . $\enspace$ Be $\enspace k\in\mathbb{N}$ .
$(2)\hspace{5mm} \displaystyle\int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x} = \frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2}x+k\pi)^2 \sin(\frac{\pi}{2}x)}+ $$\displaystyle\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2}x+(k+\frac{1}{2})\pi)^2\cos(\frac{\pi}{2}x)}$
Using $\enspace \cos(\frac{\pi}{2}x)\geq 1-x\enspace $ and $\enspace \sin(\frac{\pi}{2}x)\geq x\enspace $ for $\enspace 0\leq x\leq 1\enspace$
and because of $\enspace\displaystyle 1+(\frac{\pi}{2})^2(x+2k)^2 x \leq 1+(\frac{\pi}{2})^2(2-x+2k)^2 x \enspace$ we get
$(3)\hspace{5mm} \displaystyle\int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x} \leq $$\displaystyle\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(x+2k)^2 x}+\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(x+2k+1)^2 (1-x)} $
$\hspace{2cm}\displaystyle =\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(x+2k)^2 x} +\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(2-x+2k)^2 x} $
$\hspace{2cm}\displaystyle\leq \pi\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(x+2k)^2 x} \leq \pi\int\limits_0^1\frac{dx}{1+\pi^2 kx(k+x)} $
$(4)\hspace{5mm} \displaystyle\int\limits_0^1 \frac{dx}{1+\pi^2 kx(k+x)} = \frac{1}{ \sqrt{\pi^2 k(\pi^2 k^3-4)} }
\ln\frac{2\pi^2 kx+\pi^2 k^2-\sqrt{\pi^2 k(\pi^2 k^3-4)} }{2\pi^2 kx+\pi^2 k^2+\sqrt{\pi^2 k(\pi^2 k^3-4)} } |_0^1 $
$\hspace{2cm}\displaystyle = \frac{1}{ \pi \sqrt{k(\pi^2 k^3-4)} } \ln\frac{ 2\pi^2 k^3+4k+2\pi k\sqrt{k(\pi^2 k^3-4)} }{2\pi^2 k^3+4k-2\pi k\sqrt{k(\pi^2 k^3-4)}} \leq \frac{1}{\pi\sqrt{\pi^2-4}}\frac{\ln(1+\pi^2 k^2)}{k^2}$
$\hspace{2cm}\displaystyle\leq \frac{1}{\pi\sqrt{\pi^2-4}}\frac{\ln((1+\pi^2) k^2)}{k^2}=
\frac{\ln(1+\pi^2)}{\pi\sqrt{\pi^2-4}}\frac{1}{k^2}+\frac{2}{\pi\sqrt{\pi^2-4}}\frac{\ln k}{k^2}$
Therefore we get an upper bound for the integral, an estimation is:
$$1.5<\int\limits_0^\infty\frac{dx}{1+x^2 |\sin x|} < 1.52+\frac{\ln(1+\pi^2)}{\sqrt{\pi^2-4}} \zeta(2) - \frac{2}{\sqrt{\pi^2-4}}\zeta’(2)$$
Note:
It's $\enspace\displaystyle\zeta(2)=\frac{\pi^2}{6}\enspace$ and
$\enspace\displaystyle -\zeta'(2)=\sum\limits_{k=1}^\infty\frac{\ln k}{k^2}\approx 0.93755 \enspace$ .