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.
