2个回答
展开全部
1- sinx
=1- 2sin(x/2).cos(x/2)
= [sin(x/2)]^2 -2sin(x/2).cos(x/2) + [cos(x/2)]^2
=[sin(x/2) - cos(x/2) ]^2
∫(0->π) √(1-sinx) dx
=∫(0->π) |sin(x/2) - cos(x/2) |dx
=∫(0->π/2) [cos(x/2) - sin(x/2) ]dx + ∫(π/2->π) [sin(x/2) - cos(x/2) ]dx
=4(√2-1)
=1- 2sin(x/2).cos(x/2)
= [sin(x/2)]^2 -2sin(x/2).cos(x/2) + [cos(x/2)]^2
=[sin(x/2) - cos(x/2) ]^2
∫(0->π) √(1-sinx) dx
=∫(0->π) |sin(x/2) - cos(x/2) |dx
=∫(0->π/2) [cos(x/2) - sin(x/2) ]dx + ∫(π/2->π) [sin(x/2) - cos(x/2) ]dx
=4(√2-1)
追问
非常感谢!!^_^
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