# What is the integral of 1/(5+cosx) ?

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We'll apply the following strategy of solving: we'll use the substitution tan (x/2) = t => x = 2 arctan t.

We'll differentiate both sides:

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

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

Int dx/(5+cosx) = Int 2dt/[(5 + (1 - t^2)/(1+t^2))*(1+t^2)]

Int 2dt/[(5 + (1 - t^2)/(1+t^2))*(1+t^2)] = 2 Int dt/(5+5t^2+1-t^2)

2 Int dt/(5+5t^2+1-t^2) = 2Int dt/(4t^2 + 6)

2Int dt/(4t^2 + 6) = 2Int dt/4(t^2 + 6/4)

(1/2)*Int dt/(t^2 + 3/2) = sqrt2/2sqrt3)arctan(t*sqrt6/3) + C

**Int dx/(5+cosx) = (sqrt6/6)arctan(tan(x/2)*sqrt6/3) + C**