令x-u=t,du=-dt
u=0,t=x
u=x,t=0
∫[0,x] f(x-u)e^udu
=∫[x,0] f(t)e^(x-t)*(-dt)
=∫[0,x] f(t)e^(x-t)dt
=e^x∫[0,x] f(t)e^(-t)dt
=sinx
∫[0,x] f(t)e^(-t)dt=sinx/e^x
两边求导得
f(x)e^(-x)=(sinx/e^x)'
=(cosxe^x-sinxe^x)/(e^x)^2
=(cosx-sinx)/e^x
f(x)=cosx-sinx