Line integral evaluation 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 $t\in \mathbb{R}, 0<t<2\pi$, and direction of $C$ is the same as the direction of growth of variable $t$. 

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 :)
 A: Why "this is ugly"?
Following your first step, you only need to find:
$$
\int \sin^2 x,
\int \cos^2x\sin x,
\int \cos^3x-\sin^3x,
\int \sin x\cos x, 
\int \cos^3x\sin x,
\int \sin^3x\cos x
$$
one by one. 
Can you use $\int \cos^2 xd(\cos x)$ for one of them? You might also want to change $\sin^2x$ into a function of $\cos 2x$.
A: Since sin and cos are periodic, shift the domain of $\theta$ from $[0,2 \pi]$ to $[- \pi, \pi]$. Since sin is odd, $\int \limits_{-\pi}^\pi \cos^2 t \sin t dt $ , $\int \limits_{-\pi}^\pi \sin^3 t dt $ , $\int \limits_{-\pi}^\pi \cos t \sin t dt $ , $\int \limits_{-\pi}^\pi \cos^3 t \sin t dt $ and $\int \limits_{-\pi}^\pi \cos t \sin^3 t dt $ all vanish, so only $\int \limits_0^{2\pi} \cos^3 t dt $ and $\int \limits_0^{2\pi} \cos^2 t dt$ remain.
Shifting the domain from $[0,2 \pi]$ to $[- {\pi \over 2}, {3\pi \over 2}]$,
$$
\int \limits_0^{2\pi} \cos^3 t dt
= \int \limits_{-\pi \over 2}^{3\pi \over 2} \sin^3 ({\pi \over 2} - t) dt
= \int \limits_{-\pi}^\pi \sin^3 u du
= 0
$$
Using the same trick,
$
\int \limits_0^{2\pi} \cos^2 t dt = \int \limits_0^{2\pi} \sin^2 t dt
$.
Hence
$$
\int \limits_0^{2\pi} \cos^2 t dt = {1 \over 2} \int \limits_0^{2\pi} (\sin^2 t + \cos^2 t) dt = \pi
$$
Therefore, answer is $-a^2 \pi$.
Another way to simplify the integral is to find a potential for the field.
Let $F = (y-z,x^2-y,z-x)$, $F_1 = (y,x^2,0)$ and $f = xz-{y^2 \over 2} -{z^2 \over 2}$. Then $F = \nabla f + F_1$.
Therefore, $$\oint \limits_C \langle F,T \rangle ds = \oint \limits_C \langle F_1,T \rangle ds + 0 = \oint \limits_C \langle F_1,T \rangle ds $$
which gives the same answer.
A: I would continue (1):
$a$ is a constant, doesn't bother, and there are lots to know about $\cos$ and $\sin$. First of all, $\cos 2t$ is fine as it is. $\sin^3t$ will be related to $\cos 3t$ and $\sin 3t$...
