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