Find line integral:
$I = \oint\limits_{C} (y-z)\mathrm{d}x + (x^2-y)\mathrm{d}y + (z-x)\mathrm{d}z$
where curve C is given with: $\begin{array}\\ x = a\cos{t}\\ y = a\sin{t}\\ z = a^2\cos{2t}\\ \end{array}$
and where $t\in \mathbb{R}, 0
What I've tried so far:
1) If we go standardly: $\mathrm{d}x=-a\sin{t}\mathrm{d}t$, $\mathrm{d}y=a\cos(t)\mathrm{d}t$, $\mathrm{d}z=-4a^2\sin{t}\cos{t}\mathrm{d}t$, and substitute all into integral, we get: $\int\limits_{0}^{2\pi}-a^2\sin^2{t} + 3a^3\cos^2{t}\sin{t} - a^3\sin^3{t} + a^3\cos^3{t} - a^2\sin{t}\cos{t}-4a^4\cos^3{t}\sin{t} + 4a^4\sin^3{t}\cos{t})\mathrm{d}t$
but this is very ugly and I have no idea how to proceed, apart from trying random trig manipulations, which I've tried to no success.
2) If we realize that $z = a^2\cos^2{t} - a^2\sin^2{t} = x^2 - y^2$, and that $\mathrm{d}x=-y\mathrm{d}t$, $\mathrm{d}y=x\mathrm{d}t$, and $\mathrm{d}z=-4xy\mathrm{d}t$, we can substitute that into the integral, and get: $ \int\limits_0^{2\pi}\left(y^2+3x^2y-y^3+x^3-xy-4x^3y+4xy^3\right)\mathrm{d}t, $ but this is also kind of hopeless :)