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

Question

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.

Answer 1

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. (>‿◠)✌

Answer 2

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.

Answer 3

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 !!).

Answer 4

I would like to try to tackle the problem with elementary integration techniques though it is a bit tedious. First of all, using double angle formula, we get $$\cos x\cos 4x= 8 \cos ^{5} x-8 \cos ^{3} x+\cos x$$ and hence $$ I=\int_{0}^{\pi} \frac{8 \cos ^{5} x-8 \cos ^{3} x+\cos x}{(2-\cos x)^{2}} d x $$

By division, we reduce the integrand to a proper fraction, $$ \begin{aligned} I \displaystyle &=\int_{0}^{\pi}\left[8 \cos ^{3} x+32 \cos ^{2} x+88 \cos x+224\right] d x + \displaystyle \int_{0}^{\pi} \frac{545 \cos x-896}{(\cos x-2)^2} d x \\&= 240 \pi+545\underbrace{\int_{0}^{\pi} \frac{d x}{\cos x-2}}_{J}+194 \underbrace{\int_{0}^{\pi} \frac{d x}{(\cos x-2)^{2}}}_{K} \end{aligned} $$

For any $a>1,$ let’s define a definite integral$$ \begin{aligned} \displaystyle I(a):&=\int_{0}^{\pi} \frac{d x}{a-\cos x} \stackrel{x\mapsto \pi-x}{=} \int_{0}^{\pi} \frac{d x}{a+\cos x}\\ \\2 I(a)&=\int_{0}^{\pi}\left(\frac{1}{a-\cos x}+\frac{1}{a+\cos x}\right) d x\\ &=4 a \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2}-\cos ^{2} x} d x\\ &=4 a \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{a^{2} \sec ^{2} x-1} d x\\ &=4 a \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{a^{2} \tan ^{2} x+\left(a^{2}-1\right)}\\ &=\frac{4}{\sqrt{a^{2}-1}}\left[\tan ^{-1}\left(\frac{a \tan x}{\sqrt{a^{2}-1}}\right)\right]_{0}^{\frac{\pi}{2}}\\ &=\frac{2\pi}{\sqrt{a^{2}-1}}\\ \therefore \quad \int_{0}^{\pi} \frac{d x}{a-\cos x}&=\frac{\pi}{\sqrt{a^{2}-1}} \end{aligned} $$ In particular, $$J=-I(2)=-\frac{\pi}{\sqrt{3}}$$ For $K$, we just differentiate $I(a)$ w.r.t. $a$.

$$ \begin{aligned} I^{\prime}(a)=-\frac{\pi}{\left(a^{2}-1\right)^{\frac{3}{2}}} &\Rightarrow \int_0^{\pi}\frac{1}{(a-\cos x)^{2}} d x=\frac{\pi a}{\left(a^{2}-1\right)^{\frac{3}{2}}} \\ \therefore K=\int_{0}^{\pi} \frac{1}{(2-\cos x)^{2}} d x&=\frac{2 \pi}{3 \sqrt{3}} \\ \end{aligned} $$

Now we can conclude that

$$I =240 \pi+545\left(-\frac{\pi}{\sqrt{3}}\right)+194\left(\frac{2 \pi}{3 \sqrt{3}}\right) =\frac{\pi}{9}(2160 - 1247 \sqrt{3})$$

Answer 5

Contour Integration

We first split the integral into two $$I= \int_0^\pi \frac{\cos x \cos 4 x}{(2-\cos x)^2} d x =2 \int_0^\pi \frac{\cos 4 x}{(2-\cos x)^2} d x-\int_0^\pi \frac{\cos 4 x}{2-\cos x}dx$$

Now let $$I(a)= \int_0^{2\pi }\frac{\cos 4 x}{a-\cos x} dx $$ where $|a|>1.$

Then evaluating $I(a)$ via contour integration along the unit circle $|z|=1$ in anti-clockwise direction yields Then use the property of cosine to convert $$ \begin{aligned} I(a) & =\frac{1}{2} \int_0^{2 \pi} \frac{\cos 4 x}{a-\cos x} d x \\ & =\frac{1}{2} \operatorname{Re} \oint_{|z|=1} \frac{z^4}{a-\frac{1}{2}\left(z+\frac{1}{z}\right)} \frac{d z}{i z}\\&= -\operatorname{Re}\left(\frac{1}{i} \oint_{|z|=1} \frac{z^4}{\left[z-\left(a+\sqrt{a^2-1} \right)\right]\left[z-\left(a-\sqrt{a^2-1}\right)\right]}\right) \\\\ \end{aligned} $$

We now have a simple pole $z=a-\sqrt{a^2-1}$ within the unit circle. By Cauchy Residue Theorem, we have $$ \begin{aligned} I(a) & =-\operatorname{Re}\left(2 \pi \cdot \frac{\left(a-\sqrt{a^2}-1\right)^4}{\left(a-\sqrt{a^2-1}\right)-\left(a+\sqrt{a^2-1}\right)}\right) \\ & =\frac{\pi}{\sqrt{a^2-1}}\left(a-\sqrt{a^2-1}\right)^4 \end{aligned} $$

$$ \begin{aligned} \boxed{\int_0^\pi \frac{\cos 4 x}{2-\cos x} =\frac{\pi}{\sqrt{3}}(2-\sqrt{3})^4 =\frac{\pi}{3}(97 \sqrt{3}-168)} \end{aligned} $$

Differentiating $I(a)$ at $a=2$ yields

$$ \begin{aligned} -\int_0^\pi \frac{\cos 4 x}{(2-\cos x)^2} d x & =\left.\pi \frac{d}{d a} \frac{\left(a-\sqrt{a^2-1}\right)^4}{\sqrt{a^2-1}}\right|_{a=2} \\ & =-\left.\pi \frac{\left(a-\sqrt{a^2-1}\right)^4\left(4 \sqrt{a^2-1}+a\right)}{\left(a^2-1\right)^2}\right|_{a=2}\\ \end{aligned} $$ Hence $$ \begin{aligned} \boxed{\int_0^\pi \frac{\cos 4 x}{(2-\cos x)^2} d x =\pi \frac{(2-\sqrt{3})^4(4 \sqrt{3}+2)}{3 \sqrt{3}} =\frac{\pi}{9}(828-478 \sqrt{3})} \end{aligned} $$ $$ \begin{aligned} I & =2 \int_0^\pi \frac{\cos 4 x}{(2-\cos x)^2} d x-\int_0^\pi \frac{\cos 4 x}{2-\cos x} d x \\ & =\frac{2 \pi}{9}(828-478 \sqrt{3})-\frac{\pi}{3}(97 \sqrt{3}-168) \\ & =\boxed{\frac{\pi}{9}(2160-1247 \sqrt{3})} \end{aligned} $$