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

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### 2 Answers

We have to find the anti-derivative of [ cos 2x - (cos x)^2]^-1

f(x) = [ cos 2x - (cos x)^2]^-1

=> f(x) = [(cos x)^2 - (sin x)^2 - (cos x)^2]^-1

=> f(x) = -(sin x)^-2

=> f(x) = -1 / (cosec x)^2

Int [ -1 / (cosec x)^2 dx] = cot x + C

The required anti-derivative is **cot x + C**

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**