1个回答
展开全部
设x=2u
∫(0,π/2)(x+sinx)/(1+cosx)dx
=∫(0,π/2)(2u+sin2u)/(1+cos2u)d(2u)
=2∫(0,π/4)(2u+2sinucosu)/(1+2cos²u-1)du
=2∫(0,π/4)(u+sinucosu)/cos²udu
=2∫(0,π/4)u/cos²udu+2∫(0,π/4)sinu/cosudu
=2∫(0,π/4)udtanu+2∫(0,π/4)sinu/cosudu
=2utanu-2∫(0,π/4)tanudu+2∫(0,π/4)sinu/cosudu
=2utanu|(0,π/4)
=xtan(x/2)|(0,π/2)
=π/2
∫(0,π/2)(x+sinx)/(1+cosx)dx
=∫(0,π/2)(2u+sin2u)/(1+cos2u)d(2u)
=2∫(0,π/4)(2u+2sinucosu)/(1+2cos²u-1)du
=2∫(0,π/4)(u+sinucosu)/cos²udu
=2∫(0,π/4)u/cos²udu+2∫(0,π/4)sinu/cosudu
=2∫(0,π/4)udtanu+2∫(0,π/4)sinu/cosudu
=2utanu-2∫(0,π/4)tanudu+2∫(0,π/4)sinu/cosudu
=2utanu|(0,π/4)
=xtan(x/2)|(0,π/2)
=π/2
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
