求不定积分：∫sin[x^(1/2)]dx

求不定积分：∫sin[x^(1/2)]dx

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
=…………

如下

dx = (3/2)∫ sin(2x/3) d(2x/3) = (-3/2)cos(2x/3) + c ∫ e^sinx * cosx dx = ∫ e^sinx dsinx = e^sinx + c ∫ 1/x

dx = (3/2)∫ sin(2x/3) d(2x/3) = (-3/2)cos(2x/3) + c ∫ e^sinx * cosx dx = ∫ e^sinx dsinx = e^sinx + c ∫ 1/x