let
u=2π-x
du=-dx
I
=∫(0->2π) (x-sinx)(1-cosx)^2 dx
=∫(2π->0) (2π-u+sinu)(1-cosu)^2 (-du)
=∫(0->2π) (2π-x+sinx)(1-cosx)^2 dx
2I
=∫(0->2π) (x-sinx)(1-cosx)^2 dx +∫(0->2π) (2π-x+sinx)(1-cosx)^2 dx
=2π∫(0->2π)(1-cosx)^2 dx
=2π∫(0->2π) [1-2cosx+(cosx)^2] dx
=π∫(0->2π) [3-4cosx+cos2x] dx
=π[3x-4sinx+(1/2)sin2x]|(0->2π)
=6π^2
I=3π^2
ie
∫(0->2π) (x-sinx)(1-cosx)^2 dx =3π^2
u=2π-x
du=-dx
I
=∫(0->2π) (x-sinx)(1-cosx)^2 dx
=∫(2π->0) (2π-u+sinu)(1-cosu)^2 (-du)
=∫(0->2π) (2π-x+sinx)(1-cosx)^2 dx
2I
=∫(0->2π) (x-sinx)(1-cosx)^2 dx +∫(0->2π) (2π-x+sinx)(1-cosx)^2 dx
=2π∫(0->2π)(1-cosx)^2 dx
=2π∫(0->2π) [1-2cosx+(cosx)^2] dx
=π∫(0->2π) [3-4cosx+cos2x] dx
=π[3x-4sinx+(1/2)sin2x]|(0->2π)
=6π^2
I=3π^2
ie
∫(0->2π) (x-sinx)(1-cosx)^2 dx =3π^2