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