设f(x)=(ax+b)sinx+(cx+d)cosx,若已知f'(x)=xcosx,则f(x)=
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设f(x)=(ax+b)sinx+(cx+d)cosx,若已知f''(x)=xcosx,则f(x)=设f(x)=(ax+b)sinx+(cx+d)cosx,若已知f''(x)=xcosx,则f(x)=设f
设f(x)=(ax+b)sinx+(cx+d)cosx,若已知f'(x)=xcosx,则f(x)=
设f(x)=(ax+b)sinx+(cx+d)cosx,若已知f'(x)=xcosx,则f(x)=
设f(x)=(ax+b)sinx+(cx+d)cosx,若已知f'(x)=xcosx,则f(x)=
由f(x)=(ax+b)sinx+(cx+d)cosx
知f'(x)=a*sinx+(ax+b)cosx+c*cosx-(cx+d)sinx=(-cx-d+a)sinx+(ax+b+c)cosx=xcosx
得c=0,a=d=1,b+c=0,即b=0
f(x)=xsinx+cosx
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