这三角怎么化简啊
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/08 15:09:29
这三角怎么化简啊
![](http://img.wesiedu.com/upload/d/14/d14be970944d5eaf532a21e9a56735e4.jpg)
![](http://img.wesiedu.com/upload/d/14/d14be970944d5eaf532a21e9a56735e4.jpg)
![这三角怎么化简啊](/uploads/image/z/8249876-44-6.jpg?t=%E8%BF%99%E4%B8%89%E8%A7%92%E6%80%8E%E4%B9%88%E5%8C%96%E7%AE%80%E5%95%8A%26nbsp%3B)
∫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