If dy/dx=(2+cosx)^-1 what is y ?
First, we'll re-write the expression of dy/dx, using the negative power rule:
(2+cosx)^-1 = 1/(2+cosx)
dy/dx = 1/(2+cosx) => dy = dx/(2+cosx)
To determine the primitive of the given function dy, we'll have to calculate the indefinite integral of dx/(2+cosx).
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/(2 + cos x) = 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*Int dt/[(sqrt3)^2 + t^2]
2*Int dt/[(sqrt3)^2 + t^2] = 2*sqrt3/3*arctan (tsqrt3/3) + C
But the variable t is: t = tan x/2,
The primitive of the given function is f(x) = Int dx/(cosx + 2) = (2*sqrt3/3)*arctan [(tan x/2)*sqrt3/3] + C