Partial fraction on $\frac{2\cos^2(x)-1}{a^2-2a\cos(x)+1}$

Question

$$\frac{2\cos^2(x)-1}{a^2-2a\cos(x)+1}$$ I've seen this thing on an integral on this site and this is doable with partial decomposition (also checked with Wolfram), but my question is: how? I tried factoring since what I've learned is that you have to have factors in the denominator to do the: $$\frac{A}{\text{denominator 1}}+\frac{B}{\text{denominator 2}}$$, etc. But I don't know if having a $(\sqrt{1+a^2}+\sqrt{2a\cos(x)})(\sqrt{1+a^2}-\sqrt{2a\cos(x)})$ would help to do partial decomposition? Also tried long division since the degree is higher than in the denominator but didn't work either. Any hints?

Answer 1

I suppose that you are integrating for $x$.

The first thing I would do is to use the tangent half-angle substitution $x=2 \tan ^{-1}(t)$$

$$\int\frac{2\cos^2(x)-1}{a^2-2a\cos(x)+1}\,dx=\int\frac{2 \left(t^4-6 t^2+1\right)}{\left(t^2+1\right)^2 \left((a+1)^2 t^2+(a-1)^2\right)}\,dt$$ Now $$\frac{2 \left(t^4-6 t^2+1\right)}{\left(t^2+1\right)^2 \left((a+1)^2 t^2+(a-1)^2\right)}=$$ $$-\frac{(a-1)^2}{a^2 \left(t^2+1\right)}-\frac{4}{a \left(t^2+1\right)^2}+\frac{a^4+1}{a^2 \left((a+1)^2 t^2+(a-1)^2\right)}$$