Find the indefinite integral of 1/(cos^2x-cos^4x).

First, we'll re-write the denominator. We notice that we
can factorize it by (cos x)^2:

1/[(cos x)^2 - (cos x)^4] =
1/(cos x)^2[1 - (cos x)^2]

We'll  also substitute the
numerator1, by the fundamental formula of
trigonometry:

(sin x)^2 + (cos x)^2 =
1

We notice that [1 - (cos x)^2] = (sin
x)^2

We'll re-write the
ratio:

1/(sin x)^2*(cos x)^2=[(sin x)^2 + (cos x)^2]/(sin
x)^2*(cos x)^2

1/(sin x)^2*(cos x)^2 = (sin x)^2/(sin
x)^2*(cos x)^2 + (cos x)^2/(sin x)^2*(cos x)^2

We'll
simplify the fractions:

1/(sin x)^2*(cos x)^2 = 1/(cos
x)^2 + 1/(sin x)^2

We'll integrate both
sides:

Int dx/(sin x)^2*(cos x)^2 = Int dx/(cos x)^2 + Int
dx/(sin x)^2

Int dx/(sin x)^2*(cos x)^2 = tan
x - cotan x + C