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?

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