Okay for example I want to compute a line integral along the curve described in polar coordinates by $r=\sin(2\theta)$ so I will need to parametrise this curve. (In fact I only need to parametrise one segment of the four total ones as that will allow me to figure out what I want to know.)
The problem is I don't have a clue where to start, only things I know is that $r=x^2+y^2$ and $\theta=\tan^{-1}(y/x)$ but I don't even know if that is relevant.
Similarly I don't know how to parametrise other curves given in terms of polar coordinates such as $r=1+\cos(\theta)$
Any suggestions?
Notice that $x=r\cos(\theta)$ and $y=r\sin(\theta)$, and that $\sin(2\theta)=2\cos(\theta)\sin(\theta)$.
So...
$r=\sin(2\theta)$
$r=2\cos(\theta)\sin(\theta)$
$r=\frac{2xy}{r^2}$
$r^3=2xy$
We know that $r=\sqrt{x^2+y^2}$.
So, the messy truth is that the function is described in Cartesian form by:
$(x^2+y^2)^{3} = 4x^2y^2$
However we want the parametric form of this, not the Cartesian.
Above, I mentioned that $x(\theta)=r\cos(\theta)$
We are also told that $r=\sin(2\theta)$. So we can substitution this expression for $r$ into our formula for the x-coordinate.
Consequently $x(\theta)=\sin(2\theta)\cos(\theta)$
Similarly, $y(\theta)=r \sin(\theta)$ We can substitute in the expression for $r$ into this formula to get the y-coordinate. So...
$y(\theta)=r \sin(\theta)$
$r=\sin(2\theta)$
$y(\theta)=\sin(2\theta) \sin(\theta)$
To get the parametric answer to your question, you just switch out $\theta$ for $t$ and call it a day.