解3令x= a sint dx=acst x=0→t=0x=a→1个 →a2-x2dt=[a2cos2htl 1+cos2htsm° 解4令x=ac0st 仍可得到上述结果令   − a a x dx 0 2 2  = 2 0 2 2 cos  a tdt  = + 2 0 2 (1 cos 2 ) 2  t dt a 4 2 a = 解4 令 x = acost 仍可得到上述结果 x = asin t dx = acost x = 0  t = 0 2  x = a  t = 解3