见图

来源：学生作业帮助网编辑：六六作业网时间：2024/05/25 12:50:04

见图见图见图原式=∫[0,π/2]sin(sinx)d(sinx)=[-cos(sinx)]|[0,π/2]=cos(sin0)-cos[sin(π/2)]=cos0-cos1=1-cos1

见图
见图

见图
原式=∫[0,π/2]sin(sinx)d(sinx)
=[-cos(sinx)]|[0,π/2]
=cos(sin0)-cos[sin(π/2)]
=cos0-cos1
=1-cos1

见图见图见图见图见图见图见图见图见图见图见图见图：见图见图见图见图见图见图