Assuming the particle's position is given by
[tex]\begin{cases}x(t)=3-\sin2t\\y(t)=3-\cos2t\\0\le t\le4\end{cases}[/tex]
then the distance traveled over the interval is
[tex]\displaystyle\int_{t=0}^{t=4}\sqrt{\left(\dfrac{\mathrm dx(t)}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy(t)}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^4\sqrt{(-2\cos2t)^2+(2\sin2t)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^4\sqrt{4\cos^22t+4\sin^22t}\,\mathrm dt[/tex]
[tex]=\displaystyle2\int_0^4\mathrm dt[/tex]
[tex]=2(4-0)=8[/tex]
