Astroid area with Green Theorem

Question

I know that the area of an astroid can be calculated by Green's theorem through the integral: $$ A = \frac{3a^2}{2}\int_{0}^{2\pi}[\cos^4t \sin^2t+\sin^4t\cos^2t]dt$$.

However, I found the following expression in the class notes of a former teacher of mine: $$ A = \frac{3a^2}{2}\int_{0}^{\frac{\pi}{2}}[1-\cos^2(2t)]dt$$

This integral results in the appropriate area: $\frac{3a^2 \pi}{8}$, as well as the first integral. However, I could not understand if this second integral does in fact make sense. Can you help me understand?

Answer 1

There is a match and both the answers are correct since $$ { \cos^4 x\sin^2x+\sin^4 x\cos^2x= \\\cos^2 x\sin^2x(\cos^2 x+\sin^2x)= \\\cos^2 x\sin^2x= {1\over 4}\sin^2 2x=\\ {1\over 4}(1-\cos^2 2x) } $$ Note that the integral bounds have changed to lead to a same answer.

Answer 2

With the substitution $$u=t-\pi$$

$\int_0^{2\pi}f(t)dt$ becomes $\int_{-\pi}^{\pi}f(u)du$

the integrand is an even function, so, it gives $$2\int_0^{\pi}f(u)du$$

A new substitution $ v=u-\frac{\pi}{2}$

gives $$2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}f(v)dv$$ and finally

$$4\int_0^{\frac{\pi}{2}}f(v)dv$$