Given ʃ3/[1+2cosx]
let t=tanx/2
dx=2dt/[1+t^2]
ʃ3/[1+2cosx] = 6ʃ1/[3-t^2]dt
Again....... let t=(3)^1/2tanhu
u=tanh^-1(t/3^1/2)
dt=3^1/2sech^2u
6ʃ1/[3-t^2]dt = 6(3)^1/2ʃsech^2u/[3-3tanh^2u]du
=2(3)^1/2ʃudu
=2(3)^1/2u + c
but u=tanh^-1(t/3^1/2)
=tanh^-1(tan{x/2}/3^1/2)
therefore
Given ʃ3/[1+2cosx]=tanh^-1(tan{x/2}/3^1/2) +c