Please help, how do you do this integral??

Question

Evaluate the integral:
1/(2+cosx)dx from 0 to 2pi. Help!

Answer 1

I = ∫1/(2 + cos x) dx
Now tan 2x = 2 tan x / (1 - tan² x)
tan x = 2.tan (x/2) / (1 - tan²(x/2))
let t = tan(x/2)
tan x = 2t / (1 - t²)
From this equation, a right angled triangle may be constructed in which sides are 2t, (1 - t²) with hypotenuse (1 + t²)
sinx,cosx and tanx may be readily found from this triangle.
cos x = (1 - t²) / (1 + t²)
t = tan(x / 2)
dt/dx = (1/2).sec²(x / 2)
dx = dt / [(1/2).sec²(x / 2)]
dx = 2.dt / sec²(x /2 )
I = ∫2.dt /(1 + t²)[2 + (1 - t²)/(1 + t²)]
I = ∫2.dt / [ 2t² + 2 + 1 - t²]
I = 2 ∫ dt / (t² + 3)
I = (2 /√3).[tan^(-1) (t /√3)] + C
I = (2 /√ 3). tan^(-1)[( tan(x/2) /√ 3] + C
A nightmare to type but hopefully solution is correct!
P.S.Have just noted that limits of integration of 0 to 2π are required
These give I = 0 (makes it all worth while!)

Answer 2

the integrated equation is ln(1/(2+cosx))... because
d/dx [ln u]= u'/u.... now just evaluate it...