Antiderivative of sin ⁡ ( 1 x )I have to find the values of parameter...

Step 1
Suppose F is a primitive of f. Since f is continuous on $\mathbb{R}\setminus \{0\}$, we have
$F(y)-F(x)={\int }_x^{y}f(t)dt$
for $0<x<y$ or $x<y<0$ by the fundamental theorem of calculus. The continuity of F then forces
$F(x)=F(0)+{\int }_0^x\sin \frac{1}{t}dt$
for all $x\in \mathbb{R}$. Thus it remains to see that F is differentiable at 0 with derivative 0. Now we have
$\frac{d}{dt}(t^2\cos (t^{-1}))=\sin (t^{-1})+2t\cos (t^{-1})$
and so ${\int }_0^x\sin (t^{-1})dt={\int }_0^x\frac{d}{dt}(t^2\cos (t^{-1}))-2t\cos (t^{-1})dt=x^2\cos (x^{-1})-{\int }_0^{x}2t\cos (t^{-1})dt.$
Step 2
The first part $x^2\cos (x^{-1})$ is easily seen to be differentiable at 0 with derivative 0 there, and since
$g(t)=\{\begin{array}{ll}0& ,t=0\\ 2t\cos (t^{-1})& ,t\ne 0\end{array}$
is continuous on all of $\mathbb{R}$, the fundamental theorem of calculus asserts its differentiability on $\mathbb{R}$, and the derivative is g. Hence $F^{\prime }(0)=0-g(0)=0.$