Calculate the indefinite integral of f(x)=1/(cos2x + sin^2x)

To determine the result of the indefinite integral, 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) = cosx*cosx -
sinx*sinx

cos 2x = (cosx)^2 -
(sinx)^2

If we'll pay attention to the terms of the
denominator, we'll notive that beside cos 2x, we'll have also the term (sinx)^2. So,
we'll re-write cos 2x, with respect to the function sine
only.

We'll substitute (cosx)^2 by the difference
1-(sinx)^2:

cos 2x = 1-(sinx)^2 -
(sinx)^2

cos 2x =
1-2(sinx)^2

The denominator will
become:

cos2x + (sinx)^2 = 1-2(sinx)^2 +
(sinx)^2

cos2x + (sinx)^2 =
1-(sinx)^2

But, 1-(sinx)^2 = (cosx)^2 (from the fundamental
formula of trigonometry)

cos2x + (sinx)^2 =
(cosx)^2

The indefinite integral of f(x) will
become:

Int f(x)dx = Int dx/(cosx)^2 = tan x
+ C