What is the antiderivative (cos2x-cos^2x)^-1 ?

To determine the antiderivative, we'll have to
evaluate 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

We'll re-write the denominator, using the rule of
negative power:

 (cos2x-cos^2x)^-1 =
1/ (cos2x-cos^2x)

We notice that the terms of the
denominator are cos 2x, also the term (cosx)^2. So, we'll re-write cos 2x, with respect
to the function cosine only.

We'll substitute (sin x)^2 by
the difference 1-(cos x)^2:

cos 2x = (cosx)^2  - 1 +
(cosx)^2

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

The denominator will
become:

cos2x - (cos x)^2 =  2(cos x)^2 - 1 - (cos
x)^2

cos2x - (cos x)^2 = (cos x)^2 -
1

But, (cos x)^2 - 1 = - (sin x)^2 (from the fundamental
formula of trigonometry)

cos2x - (cos x)^2 = - (sin
x)^2

The indefinite integral of f(x) will
become:

Int f(x)dx = Int dx/- (sin x)^2 = cot
x + C