Find the area of the region bounded by the astroid, in parametric form, $(x;y)=(2cos^3 t; 2sin^3 t)$
Well, I used the formula of area given in parametric curves $\int_a^b y(t).x'(t) \,dt$. So, as it's an astroid, I know that I can find the area between $[0; \pi ]$ and multiply it by 2. Then, after differentiating and replacing, I get $2 \int_0^\pi 2sin^3(t).(-6cos^2(t)sin(t)) \,dt$
which can be written as $-24 \int_0^\pi sin^4(t).cos^2(t) \,dt$
The thing is that I integrated using trigonometric identities, but it was really tedious, and, to top it, it was wrong, because when I differentiated the result I didn't get the first function and the area was negative.