Usually, the curve is given. But not for this problem which states:
a particle of mass m moves under the influence of the force field
$ F = a \sin \omega t i, a \cos \omega t j $
prove that work done up time t is $ a^2 / m \omega^2 (1 - \cos \omega t) $
It looks like a circle. So if $ r = a \cos \omega t i, a \sin \omega t j $
And $ dr = (- a \sin \omega t i, a \cos \omega t j) dt $
$work = \int F.dr $
But this line integral does not produce the result. I guess something wrong with
my curve definition.
Update.
With amd's hint the rest is easy. The idea is to compute the velocity at
time t, to get the kinetic energy of the particle, which is equal to
the work done. So first we find the acceleration by $ a = F/m $
Then we integrate acceleration to get the velocity.