# Evaluating $\int_0^{2\pi}\frac{1}{a^2+b^2-2abcos(t)}dt, 0<b<a$

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

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?

• 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 $\pi$. – Anurag A Sep 19 at 2:02
• @AnuragA This could have made a good answer in my opinion. – Allawonder Sep 19 at 4:05
• @Allawonder Thanks :-) – Anurag A Sep 19 at 18:25

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 $$d^2=\frac{c+1}{c-1}$$.