利用分部积分法
∫arcsinxarccosxdx=xarcsinxcosx-∫xd(arcsinxarccosx)
=xarcsinxcosx-∫x(arcsinxarccosx)'dx
=xarcsinxcosx-∫x[(arcsinx)'*arccosx+arcsinx*(arccosx)']dx
=xarcsinxcosx-∫x[1/√(1-x^2)*arccosx+arcsinx*[-1/√(1-x^2)]]dx
=xarcsinxcosx-∫x(arccosx-arcsinx)/√(1-x^2)dx
追问麻烦再问您下。打圆圈的这个符号是怎么变的?还有一个x去哪里了
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/cc11728b4710b9121f1c2f15c6fdfc0393452278?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
追答d(√(1-x^2))=-x/√(1-x^2)dx