例2计算曲线y= %nmn60的弧长(0sx≤mm) 解y =n sin SIn n s=、1+ 2 nTC tsin =dx x= nt ri √1+sint.ndt 0 2 SIn cos -+2sin-cos-dt 22 T n sin -+cOS =4n 0 2例 2 计算曲线y n  d n x  = 0 sin 的弧长(0  x  n). 解 n n x y n 1  = sin  sin , n x = s y dx b a = +  2 1 dx n n x   = + 0 1 sin x = nt + t  ndt   0 1 sin dt t t t t n   +       +      = 0 2 2 2 cos 2 2sin 2 cos 2 sin dt t t n        = + 0 2 cos 2 sin = 4n