well... first complete the square of x(4-x) so it becomes 4-(x-2)2
then substitute x-2=2sin a,
x = 2sin(a) +2,
dx = 2cos(a) da
bounds:
x: 0~2,
a: -pi/2~0
anyway then sub it back in, the function to be integrated should be
∫ √[4-4(sin a)squared] da
∫ 2 (cos a)squared da
hope u get...