integrate from 0to2 3x^3 sqrt(4x^2 - x^4)dx
Let u = 4x^2 - x^4
du = 8x - 4x^3 dx
You only have 3x^3, so
-3/4 du = 3x^3 - 6x
Now we have
∫ -3/4 √u du + ∫6x √(4x^2-x^4) dx
Now let's work with
∫6x √(4x^2-x^4) dx = ∫6x √(4-(x^2-2)^2) dx
If we let v = x^2-2
dv = 2x dx
we end up with
3∫√(4-v^2) dv
You can do that with integration by parts.
If you get stuck, come back with some work to show how far you got.