Find integral $$\int_C(2z^2+8z+1)dz$$ where $C$ is the parametric curve $$x = a (\theta- \sin\theta)$$ $$y = a (1 - \cos\theta)$$ $$0 \le\theta\le 2\pi$$
What i've been taught it to make the whole integral in $\theta$. So basically to get this: $$\int^{2\pi}_0f(\theta)d\theta$$ I've so far gotten $$z = a(\theta - \sin\theta + i - i\cos\theta)$$ $$\frac{dz}{d\theta}= a(1-\cos\theta +i\sin\theta)$$
My only problem is that there will be quite a lot of work to change $2z^2+8z+1$ in terms of $\theta$. So i'm just checking if what i'm doing is right or is there a different method i should be looking at.
Hint. Let $F(z):=\frac{2z^3}{3}+4z^2+z$ then $\frac{dF}{dz}(z)=f(z):=2z^2+8z+1$. Therefore \begin{align}\int_Cf(z)dz&=\int_0^{2\pi}f(z(\theta))\frac{dz}{d\theta}d\theta =\int_0^{2\pi}\frac{dF}{dz}(z(\theta))\frac{dz}{d\theta} d\theta\\ &=\int_0^{2\pi}\frac{d(F(z(\theta))}{d\theta}d\theta=F(z(2\pi))-F(z(0)) =F(2\pi a)-F(0). \end{align}