let f(x) = xsinx/(1+(cosx)^2
f(-x) = f(x)
ie f(x) is even function
∫(0,π）xsinx/(1+(cosx)^2) dx = ∫(-π,0）xsinx/(1+(cosx)^2) dx
I= ∫(0,π）xsinx/(1+(cosx)^2) dx (1)
let y =(π-x)
dy = dx
x=0,y=π
x=π ,y=0
I= ∫(π,0）(π-y)(siny) /(1+(cosy)^2) (-dy)
= -∫(0,π）(π-y)siny/(1+(cosy)^2)dy
= -∫(0,π）(π-x)sinx/(1+(cosx)^2)dx
I= -∫(0,π）πsinx/(1+(cosx)^2)dx + I
=> ∫(0,π）πsinx/(1+(cosx)^2)dx =0
let y' =-(π-x)
dy' = dx
x=0,y'=-π
x=π ,y' =0
I= ∫(-π,0）-(π-y')(-siny') /(1+(cosy')^2) (dy')
= ∫(-π,0）(π-y')siny/(1+(cosy')^2)dy'
= ∫(-π,0）(π-x)sinx/(1+(cosx)^2)dx
= ∫(-π,0）πsinx/(1+(cosx)^2)dx -I
2I = ∫(-π,0）πsinx/(1+(cosx)^2)dx
let g(x) = πsinx/(1+(cosx)^2)dx
g(-x) = -g(x) ( g is odd)
∫(-π,0）πsinx/(1+(cosx)^2)dx = -∫(0,π）πsinx/(1+(cosx)^2)dx =0
=> I= ∫(0,π）xsinx/(1+(cosx)^2) dx =0
