pure integration

∫ 1/(x^2 + 2cosαx + 1) dx

= ∫ 1/[(x + cosα)^2 + (1 - cos^2α)] dx

= ∫ 1/[(x + cosα)^2 + (sin^2α)] dx

Let y = x + cosα. When x = 0, y = cosα. When x = 1, y = 1 + cosα

So ∫ 1/(x^2 + 2cosαx + 1) dx

= ∫ 1/[y^2 + (sin^2α)] dy [from cosα to 1 + cosα]

= [1/sinα]arctan(y/sinα) |[cosα, 1 + cosα]

= [1/sinα]{arctan[(1 + cosα)/sinα] - arctan(cosα/sinα)}

= [1/sinα]arctan{[(1 + cosα)/sinα - (cosα/sinα)]/[(1 + cosα)/sin^2α]}

= [1/sinα]arctan[sinα/(1 + cosα)]

= [1/sinα]arctan[2sin(α/2)cos(α/2)/2cos^2(α/2)]

= (1/sinα)(α/2)

= α/(2sinα)