求不定积分:∫sin[x^(1/2)]dx
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:2024/07/11 06:47:30
求不定积分:∫sin[x^(1/2)]dx
![求不定积分:∫sin[x^(1/2)]dx](/uploads/image/z/7536780-36-0.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%EF%BC%9A%E2%88%ABsin%5Bx%5E%281%2F2%29%5Ddx)
原积分=∫sin[x^(1/2)]×2x^1/2dx^1/2,
令x^1/2=t,则原式=∫sint×2tdt
=﹣2∫tdcost
=﹣2tcost+2∫costdt
=﹣2tcost+2sint+C
=…………
令x^1/2=t,则原式=∫sint×2tdt
=﹣2∫tdcost
=﹣2tcost+2∫costdt
=﹣2tcost+2sint+C
=…………