# If f'(x)=1/(7+cosx) what is f(x) ?

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To determine the primitive of the given function f'(x), we'll have to calculate the indefinite integral of f'(x).

Int f'(x)dx = Int dx/(7+cos x)

This is a trigonometric integral and we'll turn it into an integral of a rational function. We'll replace tan (x/2) by the variable 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/(7 + cos x) = Int [2dt/(1 + t^2)]/[7 + (1-t^2)/(1+t^2)]

Int [2dt/(1 + t^2)]/[(7 + 7t^2 + 1 - t^2)/(1+t^2)]

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

Int 2dt/(8 + 6t^2) = Int 2dt/2(4 + 3t^2)

Int 2dt/2(4 + 3t^2) = Int dt/3(4/3 + t^2)

Int dt/3(4/3 + t^2) = (1/3)*Int dt/[(2/sqrt3)^2 + t^2]

(1/3)*Int dt/[(2/sqrt3)^2 + t^2] = (1/3)*sqrt3/2*arctan (tsqrt3/2) + C

But the variable t is: t = tan x/2,

**The primitive of the given function is f(x) = Int dx/(cosx + 7) = (sqrt3/6)*arctan [(tan x/2)*sqrt3/2] + C**