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

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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**