$\int_C{xdy-ydx}$; where $C$ is the curve $x=a\cos^3 t$, $y=a\cos^3t$ When I do whatever work for the working,i.e replacing the x,y,dx and dy in terms of t, I get
$3a^2\int{(\cos^2t \sin^2t})dt$.
How do I compute this integral.I would like to do the Riemann integral here, but I do not know the limits
For a Riemann sum
$S_n=\frac{b-a}{n}.∑_{k=0}^{k=n}.f\left(a+k\left(\frac{b-a}{n}\right)\right)$;
Would not my $∑_{k=0}^{k=n}f\left(a+k(\frac{b-a}{n})\right)$ become 0 here because sin and cos are complementary?
Starting from
$3a^2\int_0^{2\pi}{(\cos^2t \sin^2t})dt$
=$3a^2\int_0^{2\pi}{(\cos t \sin t})^2dt$
=$\frac{3a^2}{4}\int_0^{2\pi}{(\cos t \sin t})^2dt$
=$\frac{3a^2}{4}\int_0^{2\pi}(\sin 2t)^2dt$
$\cos 4t=1-2\sin^2 2t$ $\sin^2 2t=\frac{1}{2}(1-4\cos 4t)$
=$\frac{3a^2}{4}\int_0^{2\pi}\frac{1}{2}(1-4\cos 4t)dt$
=$\frac{3a^2}{8}\int_0^{2\pi}(1-4\cos 4t)dt$
=$\frac{3a^2}{8}[t-\sin 4t]_0^{2\pi}$
=$\frac{3\pi a^2}{4}$