i need help solving the integral x^3 sqrt[4x^2-x^4] dx with limits from 0 to 2...should i use a trig substitution formula to solve? the cosine formula? i'm a bit confused. how would i do that exactly?

2. x^3 sqrt[4x^2-x^4] dx = x^4 sqrt[4-x^2] dx

x = 2 cost
dx = 2 sint dt

hence 64(cos^4t)(sin^2t)dt

and you can proceed from here..