Given the vector field $F = (6x + 4y^3, 12x^2)$ compute the line integral $$\int_\gamma F \cdot \mathrm dP,$$ where $$\gamma\,\colon \begin{cases} x = t - t^3\\ y = \sin^2(2\pi t) \end{cases}\qquad t \in [0, 1].$$
I went through the usual route $$\int_\gamma F \cdot \mathrm dP = \int_0^1 F(\gamma(t)) \cdot \gamma'(t)\,\mathrm dt$$ but after expanding the scalar product the expression becomes really unwieldy and difficult to integrate. The last term is, for example, $-24\pi t^3\sin^4(2\pi t)\sin(4\pi t)$!?
So I suspected this might be an application of Green's theorem. I already verified that the curve is closed since $\gamma(0) = \gamma(1)$, but I wouldn't know how to integrate over the area enclosed by $\gamma$. Is there a way to apply the theorem?
Green's theorem doesn't apply directly since, as per wolfram alpha plot, $\gamma$ is has a self-intersection, i.e. is not a simple closed curve.
Also, going by the $-24\pi t^3\sin^4(2\pi t)\sin(4\pi t)$ term you mentioned, I get a different (but still awful) scalar expansion:
$$\vec F=(6x+4y^3,12x^2), x=t-t^3, y=sin^2(2\pi t),\\ x'(t)=1-3t^2,\\ y'(t)=2\sin(2\pi t)\cos(2\pi t)2\pi=2\pi\sin(4\pi t)\\ \int_\gamma \vec F\cdot dP = \int_\gamma F_1\dfrac{dx}{dt}+F_2\dfrac{dy}{dt}~dt\\ =\int_0^1\left[6(t-t^3)+4\sin^6(2\pi t)\right](1-3t^2)+12(t-t^3)^2(2\pi\sin(4\pi t))~dt\\ =\int_0^1 18t^5-24t^3+6t+4\sin^6(2\pi t)-12t^2\sin^6(2\pi t)~dt\\ -24\pi\int_0^1 t^6\sin(4\pi t)-2t^4\sin(4\pi t)+t^2\sin(4\pi t)~dt $$