设f(x)=s(1,x）cos(t^2)dt.求设s(0,1)f(x)dx.望赐教!

f '(x)=cos(x^2)
∫[0-->1] f(x)dx=[0-->1] xf(x)-∫[0-->1] xf '(x)dx
=f(1)-∫[0-->1] xcos(x^2)dx 注意f(1)=0
=-1/2∫[0-->1] cos(x^2)d(x^2)
=-1/2sin(x^2) [0-->1]
=-1/2sin1