作业帮这三角怎么化简啊
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/04/29 00:24:16
这三角怎么化简啊
∫1/(1+sinx)dx
=∫[(1-sinx)/(1-sinx)(1+sinx)]dx
=∫(1-sinx)/(1-sin^2 x)dx
=∫[(1-sinx)/cos^2 x]dx
=∫[1/cos^2 x]dx+∫[1/cos^2 x]d(cosx)
=∫[(sin^2 x+cos^2 x)/cos^2 x]dx-(1/cosx)
=∫[sin^2 x/cos^2 x]dx+x-(1/cosx)
=-∫[sinxd(cosx)]/cos^2 x+x-(1/cosx)
=∫sinxd[1/cosx]+x-(1/cosx)
=(sinx/cosx)-∫[1/cosx]cosxdx+x-(1/cosx)
=tanx-x+x-(1/cosx)+C
=(sinx-1)/cosx+C
请采纳 O(∩_∩)O谢谢
=∫[(1-sinx)/(1-sinx)(1+sinx)]dx
=∫(1-sinx)/(1-sin^2 x)dx
=∫[(1-sinx)/cos^2 x]dx
=∫[1/cos^2 x]dx+∫[1/cos^2 x]d(cosx)
=∫[(sin^2 x+cos^2 x)/cos^2 x]dx-(1/cosx)
=∫[sin^2 x/cos^2 x]dx+x-(1/cosx)
=-∫[sinxd(cosx)]/cos^2 x+x-(1/cosx)
=∫sinxd[1/cosx]+x-(1/cosx)
=(sinx/cosx)-∫[1/cosx]cosxdx+x-(1/cosx)
=tanx-x+x-(1/cosx)+C
=(sinx-1)/cosx+C
请采纳 O(∩_∩)O谢谢