Đặt $t=\frac1x\Rightarrow dt=-\frac{1}{x^2}dx=-t^2dx\Rightarrow dx =-\frac{1}{t^2}dt$
$\int\limits\frac{1}{x^3}\sin\frac{1}{x}\cos\frac{1}{x}dx=\int t^3.\sin t \cos t\left ( -\frac{1}{t^2} \right )dt=-\frac{1}{2}\int t\sin 2tdt$
$=\frac{1}{4} \int t d(\cos 2t)=\frac{1}{4}t\cos 2t - \frac{1}{4}\int \cos 2t dt =\frac{1}{4}t\cos 2t-\frac{1}{8}\sin 2t$
$=\frac{1}{4}\frac1x\cos \frac2x-\frac{1}{8}\sin \frac2x+C.$