# If dy/dx=(cos2x+sin^2x), determine the function y?

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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)!