If dy/dx=(cos2x+sin^2x), determine the function y?
To determine the primitive,we'll have to evaluate the result of the indefinite integral.
Int dy = Int dx/[cos2x+(sin x)^2]
We'll have to re-write the denominator. We'll apply the formula of the cosine of a double angle.
cos 2x = cos(x+x)
cos 2x = cosx*cosx - sinx*sinx
cos 2x = (cosx)^2 - (sinx)^2
We notice that the terms of the denominator are cos 2x, also the term (sin x)^2. So, we'll re-write cos 2x, with respect to the function sine only.
According to Pythagorean identity, we'll substitute (cos x)^2 by the difference 1- (sin x)^2:
cos 2x = 1 - (sinx)^2 - (sinx)^2
cos 2x = 1 - 2(sinx)^2
The denominator will become:
cos2x + (sin x)^2 = 1 - 2(sinx)^2 + (sin x)^2
cos2x + (sin x)^2 = 1 - (sin x)^2
But, 1 - (sin x)^2 = (cosx)^2 (from the fundamental formula of trigonometry)
cos2x + (sin x)^2 = (cosx)^2
Int dy = Int dx/(cosx)^2 = tan x + C
Sorry, the derivative is dy/dx=1/(cos2x+sin^2x)!