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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?

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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.

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

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