I have a parametric curve (shapes like a heart):
$r(t) = (16sin^3(t), 13cos(t) - 5cos(2t) - 2cos(3t) - cos(4t))$
$0\le t \le 2\pi$
And I'm going to find the area limited by the curve.
I have to find
$\int_0^{2\pi} y*x' dt$
$\int_0^{2\pi} (13cos(t) - 5cos(2t) - 2cos(3t) - cos(4t))48sin^2(t)cos(t) dt$, which is pretty ugly.
But I'm given a hint:
$\int_0^{2\pi} sin(nx)sin(mx) dx = \int_0^{2\pi} cos(nx)cos(mx) dx = \begin{cases} \pi & \text{for m = n} \\[2ex] 0 & \text{otherwise} \end{cases}$
But I'dont know how to use this identity in this case.
hint
In your ugly integral, Replace $ \; \sin^2(t) \; $ by $ \;(1-\cos^2(t)) $
and linearise by$$\cos^3(t)=\Bigl(\frac{e^{it}+e^{-it}}{2}\Bigr)^3$$
to get $$\cos^3(t)=\frac 14\Bigl(\cos(3t)+3\cos(t)\Bigr)$$
then you can use $$\int_0^{2\pi}\cos(nt)\cos(mt)dt=\pi \; if \; m=n \; and \; 0 \; otherwise$$
Your area will be about $$180\pi$$