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We have to evaluate-

$$\int_0^\pi \frac{\sin^2x}{a^2+b^2-2ab\cos x}\mathrm dx$$

I tried applying the property $\displaystyle\int_0^{2a}f(x)\,\mathrm dx=\int_0^a f(x)+f(2a-x)$ and proceeding I got $$\frac{a^2+b^2}{2a^2b^2}\frac{\pi}{2}-\frac{2(a^2+b^2)}{4a^2b^2}\int_0^{\pi/2}\frac{(a^2-b^2)^2\sec^2x}{(a^2-b^2)^2\sec^2x+4a^2b^2\tan^2x}\mathrm dx$$

Things are getting messed up after this.

Is there a better way out?

an4s
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Soham
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    Would starting with a Weierstrass substitution yield anything helpful, perhaps? – Robert Howard Aug 07 '18 at 17:27
  • Since you are not familiar with complex numbers, I deleted my answer. Here you can find an approach using real methods. https://math.stackexchange.com/questions/1038263/calculation-of-int-0-pi-frac-sin2-xa2b2-2ab-cos-x-dx – Zacky Aug 07 '18 at 18:08
  • @Zacky I think you should undelete it... this post is not ofcourse meant only for my reference... others may find it useful... anyway thanks!! – Soham Aug 07 '18 at 18:09
  • Also this might be of use in order to find a real solution:$$ 1 + 2 \sum_{k=1}^{\infty} a^{k} \cos(kx) = 1 + 2 \left(\frac{1-a \cos x}{1-2a \cos x +a^{2}} -1\right) = \frac{1-a^{2}}{1-2 a \cos x + a^{2}}$$ – Zacky Aug 07 '18 at 18:19
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    @Zacky Thanks again... I am not into the problem at this moment... I would definitely like to get back to you if I have any doubts as soon as possible... good day! – Soham Aug 07 '18 at 18:22

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