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How may I evaluate the $$\int_0^{2\pi}\frac{1}{a^2+b^2-2ab\cos(t)}\mathrm dt, \,\,0<b<a$$?

I saw some simialr result in this site, but its integration limits are $0$ to $\pi$ and I find tough to solve it for this limit. Can it be solved by elementary method?

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    $\begingroup$ Use $f(2a-x)=f(x) \implies \int_0^{2a}f(x) \, dx=2\int_0^{a}f(x) \, dx$ to convert it from $0$ to $\pi$. $\endgroup$ – Anurag A Sep 19 at 2:02
  • $\begingroup$ @AnuragA This could have made a good answer in my opinion. $\endgroup$ – Allawonder Sep 19 at 4:05
  • $\begingroup$ @Allawonder Thanks :-) $\endgroup$ – Anurag A Sep 19 at 18:25
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Let $c=\frac{a^2+b^2}{2ab},$ then $$I=\frac{1}{2ab}\int_{0}^{2\pi} \frac{dt}{c- \cos t} = \frac{1}{ab} \int_{0}^{\pi} \frac{dt}{c-\cos t}= \frac{1}{ab} \int_{0}^{\pi} \frac{dt}{(c-1)\sin^2(t/2)+(c+1)\cos^2(t/2)}=\frac{2}{ab(c-1)}\int_{0}^{\pi/2} \frac {\mbox{sec}^2x~ dx}{\tan^2x+d^2} =\frac{2}{ab(c-1)}\int_{0}^{\infty} \frac {du}{u^2+d^2}= \frac{\pi}{ab\sqrt{c^2-1}}=\frac{2\pi}{a^2-b^2}.$$ Here $t/2=x$, $\tan x=u$ and $d^2=\frac{c+1}{c-1}$.

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