作业帮求不定积分：∫sin[x^(1/2)]dx
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/15 14:23:00
![求不定积分：∫sin[x^(1/2)]dx](/uploads/image/z/7236701-53-1.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)
x){Ɏާf=_iG= u.̋07ҌMI*'J~
&>Qa&xQE &u^ٖ< tO?T_ P}IJ X%)HB% 6HH ~ l ^s.zLJ1@fjUk*@i.Hh'TD[!Y65BAK(U 6I8L¸0M V;
求不定积分：∫sin[x^(1/2)]dx
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