解定积分

来源：百度知道 编辑：UC知道 时间：2024/06/10 19:05:38

int/elliptic: trying elliptic integration
int/ellalg/elltype: Checking for an elliptic integral (1-x^2)^(1/2)-(1/6)*x-1/6 freeof(x) x

答案: 2*sqrt(Pi)*(105/(1369*sqrt(Pi))+(1/2)*arcsin((6/37)*sqrt(37))/sqrt(Pi))-432/1369

用Maple算的，该不会错的

积分;(35/37,-1)[根号(1-x^2)-x/6-1/6]dx

积分:根号(1-x^2)dx
令x=sint,|t|<pai/2
=积分:costd(sint)
=积分:cos^2tdt
=1/2积分:(1+cos2t)dt
=1/2(t+1/2sin2t)+C
=1/2(arcsinx+x根号(1-x^2))+C

原式
=1/2(arcsinx+x根号(1-x^2)-x^2/12-x/6|(35/37,-1)
(直接代入,自己代一下,有点麻烦)