求不定积分：∫sin[x^(1/2)]dx

题目

求不定积分：∫sin[x^(1/2)]dx

答案

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
=…………