Primitive of a function with $\sin \frac{1}{x}$

I have the next integral: $$\int\biggl({\frac{\sin \frac{1}{x}}{x^2\sqrt[]{(4+3 \sin\frac{2}{x})}}}\biggr)\,dx ,\;x\in \Bigl(0,\infty\Bigr)$$ I used the substitution $$u=\frac{1}{x}$$ and I got $$-\int\biggl({\frac{\sin u}{\sqrt[]{(4+3 \sin2u)}}}\biggr)\,du$$ Can somebody give me some tips about what should I do next, please?

The substitution $$u=1/x$$ yields $$dx=-\frac{1}{u^2}\,du$$, so the integral becomes $$\int\frac{-\sin u}{\sqrt{4+3\sin2u}}\,du= \int\frac{-\sin u}{\sqrt{4+3\sin2u}}\,du$$ This can be improved by setting $$u=\pi/4-v$$, so we get $$\frac{1}{\sqrt{2}}\int\frac{\cos v-\sin v}{\sqrt{4+3\cos2v}}\,dv= \frac{1}{\sqrt{2}}\biggl( \int\frac{\cos v}{\sqrt{7-6\sin^2v}}\,dv -\int\frac{\sin v}{\sqrt{6\cos^2v+1}}\,dv \biggr)$$ that you should be able to manage.

As $$(\sin v\pm\cos v)^2=1\pm\sin2v$$

$$\int\dfrac{2\sin v\ dv}{f(\sin2v)}=\int\dfrac{(\sin v-\cos v)\ dv}{f(\sin2v)}+\int\dfrac{(\sin v+\cos v)\ dv}{f(\sin2v)}=I+J$$ where $$f(\sin2v)$$ is a function of $$\sin2v$$

As $$\displaystyle\int(\sin v-\cos v)=-(\sin v+\cos v)+C,$$ set $$\sin v+\cos v=y$$ for

$$I=\int\dfrac{(\sin v-\cos v)\ dv}{f((\sin v+\cos v)^2-1)}$$

Similarly, set $$\sin v-\cos v=z$$ for $$J=\int\dfrac{(\sin v+\cos v)\ dv}{f(1-(\sin v-\cos v)^2)}$$