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Prove that

$$\int_0^\pi\frac{\cos x \cos 4x}{(2-\cos x)^2}dx=\frac{\pi}{9} (2160 - 1247\sqrt{3})$$

I tried to use Weierstrass substitution but the term $\cos 4x$ made horrible algebraic-forms since $\cos 4x = \sin^4 x + \cos^4 x - 6\sin^2 x \cos^2 x$. My friend suggests me use a contour integration method but I am not familiar with that method. Any idea? Any help would be appreciated. Thanks in advance.

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Proposition :

\begin{equation}\int_0^\pi\frac{\cos mx}{p-q\cos x}\, dx=\frac{\pi}{\sqrt{p^2-q^2}}\left(\frac{p-\sqrt{p^2-q^2}}{q}\right)^m\qquad\hbox{for}\qquad |q|<p \end{equation}


Proof :

We have \begin{equation} \int_0^\pi\frac{\cos mx}{a^2-2ab\cos x+b^2}\, dx=\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^m\qquad\hbox{for}\qquad |b|<a\tag1 \end{equation} The complete proof is given by Prof. Omran Kouba and can be seen here.

Now, let $p=a^2+b^2$ and $q=2ab$, then $p+q=\sqrt{p+q}$ and $p-q=\sqrt{p-q}$. Therefore \begin{align} 2a&=\sqrt{p+q}+\sqrt{p-q}\\[10pt] 2b&=\sqrt{p+q}-\sqrt{p-q}\\[10pt] a^2-b^2&=\sqrt{p^2-q^2}\tag2\\[10pt] \frac{b}{a}&=\frac{p-\sqrt{p^2-q^2}}{q}\tag3 \end{align} then plugging in $(2)$ and $(3)$ to $(1)$ we prove our proposition. $\quad\square$

Set $m=4$ and $p=2$ then differentiate the proposition w.r.t. $q$ and take the limit for $q\to1$, we obtain \begin{align} \lim_{q\to1}\int_0^\pi\partial_q\left(\frac{\cos 4x}{2-q\cos x}\right)\, dx&=\lim_{q\to1}\partial_q\left(\frac{\pi}{\sqrt{4-q^2}}\left(\frac{2-\sqrt{4-q^2}}{q}\right)^4\right)\\[10pt] \int_0^\pi\frac{\cos x \cos 4x}{(2-\cos x)^2}dx&=\frac{\pi}{9} \left(\,2160 - 1247\sqrt{3}\,\right) \end{align} The last step is confirmed by Wolfram Alpha.

I think differentiating is easier than using contour integration or partial fraction decomposition. (>‿◠)✌

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  • $\begingroup$ This is a beautiful answer ! Thanks for providing it ! Cheers :-) $\endgroup$ – Claude Leibovici Oct 2 '14 at 16:23
  • $\begingroup$ @ClaudeLeibovici Your praise is exquisite! Thank you, I'm glad you enjoyed it. (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Oct 2 '14 at 16:33
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    $\begingroup$ This is a nice answer. Thank you very much. I'll wait for 2 or 3 days to give you 50 rep bounty. Right now, I upvote & accept your answer. Once again, thank you $\endgroup$ – Venus Oct 6 '14 at 8:46
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    $\begingroup$ @Venus Thanks for the bounty. Yeaayyy, this is my second bounty that I get. Once again, thank you so much... (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Oct 9 '14 at 9:34
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Contour integration is a bit less painful. For first, it is better to write our integral as: $$ I = \frac{1}{4}\int_{0}^{2\pi}\frac{\cos(3x)+\cos(5x)}{(2-\cos x)^2}\,dx, $$ then, since $\cos x = \frac{e^{ix}+e^{-ix}}{2}$, by setting $z=e^{ix}$ we get: $$ I = -\frac{i}{4}\left(\oint\frac{2(z^6+1)}{z^2(z^2-4z+1)^2}\,dz + \oint\frac{2(z^{10}+1)}{z^4(z^2-4z+1)^2}\right)$$ where the path of integration is the unit circle. Since the roots of $z^2-4z+1$ occur in $2\pm\sqrt{3}$, by computing the residues in $z=0$ and $z=2-\sqrt{3}$ (tedious but straightforward) we arrive at the wanted conclusion.

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You tried to use Weierstrass substitution but $\cdots$ you have not be patient. Doing it, I arrived to $$\int\frac{\cos x \cos 4x}{(2-\cos x)^2}dx=\int\frac{2 \left(1-t^2\right) \left(t^8-28 t^6+70 t^4-28 t^2+1\right)}{\left(t^2+1\right)^4 \left(3 t^2+1\right)^2}dt$$ Using partial fraction decomposition, the last integrand is $$-\frac{2882}{3 \left(3 t^2+1\right)}+\frac{776}{3 \left(3 t^2+1\right)^2}+\frac{320}{t^2+1}+\frac{192}{\left(t^2+1\right)^2}+\frac{64}{\left (t^2+1\right)^3}+\frac{128}{\left(t^2+1\right)^4}$$ Some of the integrals are simple. I let you the pleasure of the other.

I must confess that I would prefer something like contour integration (what I cannot do !!).

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    $\begingroup$ Why I said it's horrible because after substitution I have to decompose the integrand by hand (not machine) which is painful for me. But thanks for your answer. I appreciate it. I'm still looking for other methods. +1 $\endgroup$ – Venus Sep 30 '14 at 10:34
  • $\begingroup$ Me too ! I hope for you ! Cheers :-) $\endgroup$ – Claude Leibovici Sep 30 '14 at 10:39

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