求不定积分：∫sin[x^(1/2)]dx

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/07/07 11:41:30

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
=…………

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