Question

积分(0,pie) xsinx/(1+(cosx)^2) dx

Answer 1

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