Question

求解积分

Answer 1

sinx~x
x->0 时∫<0,sinx>sint²dt ->0
设g(x)+c=∫sint²dt
g'(u(x))=g'(u)*u'=sinu²*u'
∫<0,sinx>sint²dt=g(sinx)-g(0)
lim<x->0>(g(sinx)-g(0))/x³
=lim<x->0> sin(sinx)²cosx/3x²(罗必达法则)
=lim<x->0> [-sinxsin(sin²x)+2sinxcos²x*cos(sin²x)]/6x
=lim<x->0> [-sin(sin²x)cosx+2cos³xcos(sin²x)-6sin²xcosxcos(sin²x)-4sin²xsin(sin²x)cos³x]/6
=0/6
=0

追问

1. sinx~x
   x->0 时∫<0,sinx>sint²dt ->0
   设g(x)+c=∫sint²dt
   g'(u(x))=g'(u)*u'=sinu²*u'
   上面的u怎么来的，g'(u(x))=g'(u)*u'=sinu²*u'为什么这样？

追答

复合函数
(g(u(x)))'=g'(u)*u'
g'(x)=sinx²
即g'(sinx)=sin(sinx)²
u'=(sinx)'=cosx