integral ∫1−acos⁡x1+a2−2acos⁡xdxMy trouble: Using software, i found that if |a|&gt;1, the integral is equal toarctan⁡(x2)+1−a2|a−1||a+1|arctan⁡(|a+1||a−1|tan⁡(x2))+CIf |a|&lt;1, then the antiderivate isx−arctan⁡(x2)+1−a2|a−1||a+1|arctan⁡(|a+1||a−1|tan⁡(x2))+C

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integral ∫1−acos⁡x1+a2−2acos⁡xdxMy trouble: Using software, i found that if |a|&amp;gt;1, the integral is equal toarctan⁡(x2)+1−a2|a−1||a+1|arctan⁡(|a+1||a−1|tan⁡(x2))+CIf |a|&amp;lt;1, then the antiderivate isx−arctan⁡(x2)+1−a2|a−1||a+1|arctan⁡(|a+1||a−1|tan⁡(x2))+C

bailaretzy33 · Accepted Answer

Your antiderivative is correct for |x|2≤π2. Thus,∫02π1−acos⁡(x)a2+1−2acos⁡(x),dx=∫−ππ1−acos⁡(x)a2+1−2acos⁡(x),dx=(x2+sgn(1+a)sgn(1−a)arctan⁡(|a+1a−1|tan⁡(x2)))∣−ππ=π(1+sgn(1+a)sgn(1−a))(2π,  |a|&amp;lt;1),(0  ,|a|&amp;gt;1):

integral \int \frac{1-a\cos x}{1+a^2-2a\cos x}dx My trouble: Using software, i found

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