Definite Integral by using properties or by indefinite integration

Question

$$\int_{0}^{1}\left(2x\sin \frac{1}{x}-\cos \frac{1}{x}\right)dx$$

Am stuck in this question. Can't solve by applying any of the properties of definite integral. What should one do? By performing indefinite integration and then putting limits but that would be quite lengthy I guess.

Answer 1

Hint: set $$ f(x) = x^2\sin \frac1x $$ and find $f'(x)$.

Answer 2

Hint: Use the substitution $x\mapsto\frac1x$: $$ \begin{align} \int_0^1\left(2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)\right)\mathrm{d}x &=\int_1^\infty\frac{2\sin(x)-x\cos(x)}{x^3}\,\mathrm{d}x \end{align} $$ then notice $$ \frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin(x)}{x^2}=\frac{x\cos(x)-2\sin(x)}{x^3} $$

Answer 3

By parts,

$$\int 2x\sin\left(\frac1x\right)dx=x^2\sin\left(\frac1x\right)dx-\int -\frac{x^2}{x^2}\cos\left(\frac1x\right)dx=x^2\sin\left(\frac1x\right)+\int \cos\left(\frac1x\right)dx.$$

You can conclude from there.