积分求圆的周长。 作者：远景城 半径为1的圆的周长，可用如下积分算出： 4*Int(0,1)[1/(1-x^2)^0.5]dx = 4(arcsin(1)-arcsin(0)) arcsin(1) = 90度，arcsin(0) = 0度 所以，半径为1的圆的周长= 360。 请问错在哪里？ 解：

If x is measured in radians, then (sin x)' = cos x.
Therefore (arcsin x)' = 1/sqrt(1-x^2).

However, if x is measured in degrees, then (sin x)' = (pi/180)*cos x.
Therefore (arcsin x)' = (180/pi)*(1/sqrt(1-x^2)).
Consequently, Int 1/sqrt(1-x^2) = (pi/180)*arcsin x + C.
Now do the definite integral again!