# calculus – Astroid area with Green Theorem

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

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?