if a particle moves at time t $-\pi<t<\pi$ which is given by x(t)=sin(3t) and y(t)=2t, how do i find total distance traveled? If a particle moves at time t $-\pi&lt;t&lt;\pi$ which is given by x(t)=sin(3t) and y(t)=2t, how do i find the total distance traveled?

While I can find the displacement using pythag theorem, I need to find the total distance. If it was x'(t) and y'(t) i could integrate to find total distance but in this case it just gives me the total distance. Any ideas?
In general, the length of this curve is given by
$\int_{-\pi}^{\pi} \sqrt{(x')^{2}+(y')^{2}}dt$, which in this case leads to
$$\int_{-\pi}^{\pi} \sqrt{(x')^{2}+(y')^{2}}dt = \int_{-\pi}^{\pi} \sqrt{(-3\cos(3t))^{2}+(2)^{2}}dt = \int_{-\pi}^{\pi} \sqrt{9\cos^{2}(3t)+4}dt.$$ I don't know if you need this precise answer or a rounded value, which any computer could easily do.
Just to explain both my comment and Kavi Rama Murthy's,

The distance is equal to the arc length of the curve. I hope you can see this. Just sketch out any arbitrary curve and pretend that it's a particle (or a bumblebee!) moving around on the Cartesian plane. The arc length of the curve is the distance travelled by the particle.

The arc length is given by $\displaystyle \int ds = \int \sqrt{dx^2 + dy^2}$

Here $dx = 3\cos 3t dt$ and $dy = 2dt$, so the arc length is $\displaystyle \int_{-\pi}^{\pi}\sqrt{9\cos^2 3t + 4}dt$, where the appropriate bounds have been imposed in the last step.

The problem now is evaluating this integral. The indefinite integral has no elementary form and involves elliptic integrals. Sometimes, a definite integral can be calculated "neatly" and "cleverly" with symmetry and appropriate substitutions if the bounds are just right, but I can't see a way to do so here (and neither can Wolfram Alpha). In any case, the answer is here:

https://www.wolframalpha.com/input/?i=integrate+sqrt(9cos%5E2(3x)+%2B+4)+from+-pi+to+pi