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
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