# Evaluate the integral of f(x)=1/(2+cosx)

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This type of trigonometric integral is becoming an integral of a rational function, if we'll substitute tan (x/2) = t.

x/2 = arctan t

x = 2arctan t

We'll differentiate both sides:

dx = 2dt/(1 + t^2)

We'll write cos x = (1-t^2)/(1+t^2)

We'll re-write the integral in t:

Int dx/(2+cosx) = Int [2dt/(1 + t^2)]/[2 + (1-t^2)/(1+t^2)]

Int [2dt/(1 + t^2)]/[(2 + 2t^2 + 1 - t^2)/(1+t^2)]

We'll simplify by (1 + t^2):

Int 2dt/(3+t^2) = (2/sqrt3)*arctan (t/sqrt3) + C

But t = tan x/2,

**Int dx/(2+cosx) = (2/sqrt3)*arctan [tan (x/2)/sqrt3] + C**