A算错了，是三分之四pi

f(x) =(sinx)^3 +∫(-π->π) xf(x) dx
To find : ∫(0->π) f(x) dx
let
f(x) = (sinx)^3 + C
f(x) =(sinx)^3 +∫(-π->π) xf(x) dx
=>
C
=∫(-π->π) x[(sinx)^3 + C] dx
=∫(-π->π) x(sinx)^3 dx
=2∫(0->π) x(sinx)^3 dx
=-2∫(0->π) x(sinx)^2 dcosx
=-2∫(0->π) x[1-(cosx)^2] dcosx
=-2∫(0->π) x dcosx +2∫(0->π) x(cosx)^2 dcosx
=-2[xcosx]|(0->π) +2∫(0->π) cosx dx +(2/3)∫(0->π) x d(cosx)^3
=2π +2[sinx]|(0->π) +(2/3)[x(cosx)^3]|(0->π) - (2/3)∫(0->π) (cosx)^3 dx
=2π +0 -(2/3)π -(2/3)∫(0->π) (cosx)^2 dsinx
=(4/3)π -(2/3)∫(0->π) [1-(sinx)^2] dsinx
=(4/3)π -(2/3)[sinx-(1/3)(sinx)^3]|(0->π)
=(4/3)π
ie
f(x) = (sinx)^3 +(4/3)π
∫(0->π) f(x) dx
=∫(0->π) [(sinx)^3 +(4/3)π] dx
=(4/3)π^2 +∫(0->π) (sinx)^3 dx
=(4/3)π^2 -∫(0->π) (sinx)^2 dcosx
=(4/3)π^2 -∫(0->π) [1-(cosx)^2] dcosx
=(4/3)π^2 -[cosx-(1/3)(cosx)^3]|(0->π)
=(4/3)π^2 + 4/3