How to integrate $\int_0^{2\pi} \frac{dx}{(2+\cos(x))(3+\cos(x))}$

Question

The integral is

$$\int_0^{2\pi} \frac{dx}{(2+\cos(x))(3+\cos(x))}$$

I did try Tangent half-angle substitution but only got:

$$2\int^{2\pi}_0 \frac{dt}{9t^2-t^4-12}$$

which does not seem to be easier... am I wrong on the way? Or is there easier way?

Answer 1

Hint...You could split it first by writing $$\frac{1}{(2+c)(3+c)}=\frac{1}{2+c}-\frac{1}{3+c}$$ then use the $t=\tan\frac x2$ substitution

Note that you can use symmetry to change the integral from $\int_0^{2\pi}$ to $2\times \int_0^{\pi}$

Answer 2

I'm not sure how you arrived at:

I did try Tangent half-angle substitution but only got:

$$2\int^{2\pi}_0 \frac{dt}{9t^2-t^4-12}$$

With $t = \tan\left( \tfrac{x}{2} \right)$, you have $\cos x = \tfrac{1-t^2}{1+t^2}$ and $\tfrac{dx}{dt} = \tfrac{2}{1+t^2}$; so: $$\int \frac{\mbox{d}x}{(2+ \cos x)(3+ \cos x)} \to \int \frac{1}{\left(2+ \tfrac{1-t^2}{1+t^2}\right)\left(3+ \tfrac{1-t^2}{1+t^2}\right)}\frac{2 \, }{1+t^2}\mbox{d}t$$ Now simplifying and splitting (partial fractions): $$\int \frac{1+t^2}{\left(2+ t^2\right)\left(3+ t^2\right)}\,\mbox{d}t = \int \frac{2}{3+ t^2}\,\mbox{d}t-\int \frac{1}{2+ t^2}\,\mbox{d}t$$

Answer 3

You can see that $\cos (2\pi - x) = \cos x$ and hence the integral from $0$ to $2\pi$ is twice the integral from $0$ to $\pi$. Thus we have \begin{align} I &= 2\int_{0}^{\pi}\frac{dx}{(2 + \cos x)(3 + \cos x)}\notag\\ &= 2\int_{0}^{\pi}\frac{dx}{2 + \cos x} - 2\int_{0}^{\pi}\frac{dx}{3 + \cos x}\notag\\ &= 2\frac{\pi}{\sqrt{4 - 1}} - 2\frac{\pi}{\sqrt{9 - 1}}\notag\\ &= 2\pi\left(\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{2}}\right)\notag \end{align} The integral $$\int_{0}^{\pi}\frac{dx}{a + b\cos x}$$ for $a > |b|$ is calculated using the substitution $$(a + b\cos x)(a - b\cos y) = a^{2} - b^{2}$$ and it easily evaluates to $\pi/\sqrt{a^{2} - b^{2}}$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{2\pi}{\dd x \over \bracks{2 + \cos\pars{x}}\bracks{3 +\cos\pars{x}}} = \left.-\int_{0}^{2\pi}{\dd x \over \mu + \cos\pars{x}} \,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} \\[5mm] = &\ \left.-\int_{0}^{2\pi}{\dd x \over \mu + 1 - 2\sin^{2}\pars{x/2}} \,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} = \left.-2\int_{0}^{\pi}{\dd x \over \mu + 1 - 2\sin^{2}\pars{x}} \,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} \\[5mm] = &\ \left.-2\int_{-\pi/2}^{\pi/2}{\dd x \over \mu + 1 - 2\cos^{2}\pars{x}} \,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} = \left.-4\int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \pars{\mu + 1}\sec^{2}\pars{x} - 2}\,\dd x \,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} \\[5mm] = &\ \left.-4\int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \pars{\mu + 1}\tan^{2}\pars{x} + \mu - 1}\,\dd x \,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} \\[5mm] = &\ \left.-4\,{1 \over \mu - 1}\,\root{\mu - 1 \over \mu + 1}\int_{0}^{\pi/2} {\root{\pars{\mu + 1}/\pars{\mu - 1}}\sec^{2}\pars{x} \over \bracks{\root{\pars{\mu + 1}/\pars{\mu - 1}}\tan\pars{x}}^{2} + 1}\,\dd x \,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} \\[5mm] = &\ \left.-\,{4 \over \root{\mu^{2} - 1}}\int_{0}^{\infty} {\dd x \over x^{2} + 1}\,\dd x\,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} = \left.-\,{2\pi \over \root{\mu^{2} - 1}} \,\right\vert_{\ \mu\ =\ 2}^{\ \mu\ =\ 3} = -\,{2\pi \over \root{3^{2} - 1}} + {2\pi \over \root{2^{2} - 1}} \\[5mm] = &\ \bbx{\ds{\pars{{2\root{3} \over 3} - {\root{2} \over 2}}\pi}} \approx 1.4062 \end{align}

Answer 5

I will propose an approach that uses complex analysis. The idea is to write $\cos x= (e^{ix} + e^{-ix})/2$ and then to rewrite the integral in such a way that a parametrised contour integral appears. Then a routine application of the Residue Theorem will yield the desired result.

As stated, write:

$$I=\int_0^{2\pi} \frac{dx}{(2+\mathrm{cos}(x))(3+\mathrm{cos}(x))}=\int_0^{2\pi} \frac{dx}{\left(2 + \frac{e^{ix}+e^{-ix}}{2}\right)\left(3 + \frac{e^{ix}+e^{-ix}}{2}\right)}\\ =4\int_0^{2\pi} \frac{dx}{(4 + e^{ix} + e^{-ix})(6 + e^{ix} + e^{-ix})}\\ =4\int_0^{2\pi} \frac{(e^{ix})^2}{\left(1 + 4e^{ix} + (e^{ix})^2\right)\left(1+ 6e^{ix} +(e^{ix})^2\right)}\ dx$$

Now, through the parametrisation $z=e^{ix}$ notice that this last expression equals $-i$ times the contour integral

$$J=\oint_{|z|=1}\frac{z}{(1+ 4z + z^2)(1 + 6z + z^2)}\ dz$$

where the integral is taken counter-clockwise about the unit circle. To evaluate $J$ we need the residues in the two poles of the integrand that are contained inside the unit circle $|z|=1$, namely the poles $z_1=-2 + \sqrt{3}$ and $z_2=-3 + 2\sqrt{2}$. After a tedious calculation we obtain

$$\mathrm{Res}_{z=z_1}\ f(z)=\frac{1}{4\sqrt{3}}\\ \mathrm{Res}_{z=z_2}\ f(z)=-\frac{1}{8\sqrt{2}}$$ where $f(z)$ is the integrand of $J$.

Hence by the Residue Theorem: $$J=2 \pi i \left(\frac{1}{4\sqrt{3}} -\frac{1}{8\sqrt{2}}\right)$$ which yields the desired result: $$I=-4iJ=\pi \left(\frac{2}{\sqrt{3}} -\frac{1}{\sqrt{2}}\right).$$