I am given the following to be evaluated along a cycloid from (0,0) to ($\pi$,2)
->now the integral is this: $$\int_C{(6xy-y^2)}dx+(3x^2-2xy)dy$$
->and the path being a cycloid is this: x=$\theta$-Sin($\theta$);y=1-Cos($\theta$)
After an excruciatingly difficult session of evaluation i reached the following; $$\int_C{(6xy-y^2)}dx+(3x^2-2xy)dy=[3\theta^2-\frac{3}{2}\theta-(6\theta+4) sin(\theta)+6sin^2(\theta)-(3+3\theta^2-8\theta)cos(\theta)-\frac{3}{2}cos{\theta}sin(\theta)-\frac{1}{4}sin(2\theta)-\frac{1}{2}(\theta cos(2\theta))-sin^3(\theta)]_{\theta1}^{\theta2}$$
now my question is, how do i get my upper and lower limits? What will the limits even be? i have been trying for just these two limits but to no avail. Also; is there an easier way to evaluate such integrals? I had to do it the long way, namely by substituting the X an Y parametrized coordinates, changing differentials and evaluating each integral that popped up individually; such a time-consuming task. Is there any other method that's more suitable in these cases?
At $(0,0)$ we have $x = 0, y = 0$, so $$0 = \theta - \sin \theta\\0 = 1 - \cos \theta$$ The second equation gives $1 = \cos \theta$ whose solutions are $\theta = 2k\pi$ for integer $k$. Only for $\theta = 0$ (i.e., $k = 0$) is the first equation also true.
At $(\pi, 2)$ we have $$\pi = \theta - \sin \theta\\2 = 1 - \cos \theta$$ Now the second equation gives $\cos \theta = -1$, whose solutions are $\theta = (2k+1)\pi$ for integer $k$. And again, the first equation is also true only when $\theta = \pi$.
so your limits on $\theta$ are $0$ to $\pi$.
As for the other question, sorry, but that is the easy way. It may take a bunch of busy work, but it is straight-forward. There aren't always magical shortcuts.