How to integrate $\int x^2\sin^2(\frac{1}{x})\,\mathrm{d}x$?

Question

So I've been trying to do this integral for an hour and now I'm really stuck. Can anyone help?

Answer 1

Let $x=1/u$ to get

$$I=-\int\frac{\sin^2(u)}{u^4}~\mathrm du=\frac1{6u^3}+\int\frac{\cos(2u)}{2u^4}~\mathrm du$$

Integrate by parts to get

$$\begin{align}2I-\frac1{3u^3}&=-\frac{\cos(2u)}{3u^3}-\int\frac{2\sin(2u)}{3u^3}~\mathrm du\\ &=-\frac{\cos(2u)}{3u^3}+\frac{\sin(2u)}{3u^2}-\int\frac{2\cos(2u)}{3u^2}~\mathrm du\\&=-\frac{\cos(2u)}{3u^3}+\frac{\sin(2u)}{3u^2}+\frac{2\cos(2u)}{3u}+\int\frac{4\sin(2u)}{3u}~\mathrm du\\ &=-\frac{\cos(2u)}{3u^3}+\frac{\sin(2u)}{3u^2}+\frac{2\cos(2u)}{3u}+\frac43\operatorname{Si}(2u)+c\\ &=-\frac{x^3\cos(2/x)}{3}+\frac{x^2\sin(2/x)}{3}+\frac{2x\cos(2/x)}{3}+\frac43\operatorname{Si}(2/x)+c\end{align}$$

where $\operatorname{Si}(x)$ is the sine integral.