\[\newcommand{\C}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Res}{\operatorname{Res}} \newcommand{\Phione}{\Phi_3}\]

Goal. Prove $\int_0^{2\pi}\frac{1+2\cos t}{5+4\cos t}\,dt = 0,$ and exhibit the same zero from five vantage points:

Elementary calculus — the Weierstrass tangent half-angle substitution $u=\tan(t/2)$.

Complex analysis — the residue theorem on the unit circle $\lvert z\rvert=1$.

Number theory — the numerator is the cyclotomic polynomial $\Phi_3$ (equivalently the order-1 Dirichlet kernel); the denominator is a palindromic polynomial whose roots are a reciprocal pair.

Algebraic topology — winding numbers and the de Rham class of a meromorphic 1-form.

The Green–Stokes generalization — the residue theorem is Stokes’ theorem with punctures.

Each lens is split into a settled core (the worked computation — locked) and open threads ▸ (questions and directions, to be expanded in a later edit, in the spirit of this blog’s Direct / Indirect convention).

The claim, and a 30-second sanity check

The integrand is bounded and continuous ($5+4\cos t \ge 1 > 0$), so the integral is a finite number. Split it and use the textbook value $\int_0^{2\pi}\frac{dt}{5+4\cos t}=\frac{2\pi}{\sqrt{5^2-4^2}}=\frac{2\pi}{3}$:

\[\frac{\cos t}{5+4\cos t}=\frac14\!\left(1-\frac{5}{5+4\cos t}\right) \;\Rightarrow\; \int_0^{2\pi}\frac{\cos t}{5+4\cos t}\,dt=\frac14\!\left(2\pi-5\cdot\frac{2\pi}{3}\right)=-\frac{\pi}{3}.\]

Hence

\[\int_0^{2\pi}\frac{1+2\cos t}{5+4\cos t}\,dt=\underbrace{\frac{2\pi}{3}}_{\text{constant part}}+2\cdot\underbrace{\left(-\frac{\pi}{3}\right)}_{\cos\text{ part}}=0.\]

That settles that it is zero. The rest of this post is about why — and the why is different, and illuminating, in each of the five lenses.

A — The ladder

Rung 1 — Weierstrass: brute force that collapses

Settled core. Substitute $u=\tan(t/2)$, the rationalizing change of variable:

\[\cos t=\frac{1-u^2}{1+u^2},\qquad dt=\frac{2\,du}{1+u^2}.\]

The numerator and denominator each rationalize:

\[1+2\cos t=\frac{3-u^2}{1+u^2},\qquad 5+4\cos t=\frac{9+u^2}{1+u^2},\]

so the integrand and its measure together become a rational function on the line:

\[\frac{1+2\cos t}{5+4\cos t}\,dt=\frac{3-u^2}{9+u^2}\cdot\frac{2\,du}{1+u^2} =\frac{2(3-u^2)}{(1+u^2)(9+u^2)}\,du.\]

Partial fractions ($3-u^2=A(u^2+9)+B(u^2+1)$ gives $A=\tfrac12,\,B=-\tfrac32$) reduce it to two arctangents:

\[\int_{-\infty}^{\infty}\!\left(\frac{1}{u^2+1}-\frac{3}{u^2+9}\right)du =\Big[\arctan u-\arctan\tfrac{u}{3}\Big]_{-\infty}^{\infty} =\pi-\pi=0.\]

The antiderivative $G(u)=\arctan u-\arctan(u/3)$ even returns to the same value at both ends — the zero is visible without any cancellation lemma.

Rigor note (professor-cautious). $u=\tan(t/2)$ is singular at $t=\pi$. Cleanly, split $\int_0^{2\pi}=\int_0^{\pi}+\int_{\pi}^{2\pi}$; the first maps to $\int_0^{\infty}$, the second to $\int_{-\infty}^{0}$, and because the integrand in $u$ is even and absolutely integrable the two pieces glue to $\int_{-\infty}^{\infty}$ with no boundary contribution at the removable point.

Open threads ▸

Which whole family vanishes? For $\int_0^{2\pi}\frac{a+b\cos t}{c+d\cos t}\,dt$, the half-angle reduction gives a ratio of even rational functions; work out the exact condition on $(a,b,c,d)$ that forces $0$. (Spoiler from Rung 3: it is a statement about one Fourier coefficient.)
Why does the $u$-integrand happen to be even? Trace the evenness back to $\cos(-t)=\cos t$ and ask what breaks if a $\sin t$ term is present.

Rung 2 — Residues: the two contributions cancel

Settled core. Put $z=e^{it}$, so $dt=\frac{dz}{iz}$ and $\cos t=\frac{z+z^{-1}}{2}$. Then

\[1+2\cos t=\frac{z^2+z+1}{z},\qquad 5+4\cos t=\frac{2z^2+5z+2}{z}=\frac{(2z+1)(z+2)}{z},\]

and the $z$ in the measure and the $z$’s in numerator/denominator combine to

\[\int_0^{2\pi}\frac{1+2\cos t}{5+4\cos t}\,dt =\frac1i\oint_{\lvert z\rvert=1}\frac{z^2+z+1}{z\,(2z+1)(z+2)}\,dz.\]

Three simple poles: $z=0$, $z=-\tfrac12$ (both inside) and $z=-2$ (outside). The residue theorem (Ahlfors, 1979) now reduces the integral to the interior residues:

\[\Res_{z=0}=\left.\frac{z^2+z+1}{(2z+1)(z+2)}\right|_{z=0}=\frac{1}{1\cdot2}=+\frac12,\] \[\Res_{z=-1/2}=\left.\frac{z^2+z+1}{2z(z+2)}\right|_{z=-1/2} =\frac{\tfrac14-\tfrac12+1}{2(-\tfrac12)(\tfrac32)}=\frac{\tfrac34}{-\tfrac32}=-\frac12.\]

Lemma (residue cancellation, best paper-ready). The interior residues of the integrand sum to zero: $\Res_{z=0}+\Res_{z=-1/2}=\tfrac12-\tfrac12=0,\qquad\text{hence}\qquad \int=\frac1i\cdot 2\pi i\cdot 0=0.$

The two halves of the elementary split in the sanity check ($\tfrac{2\pi}{3}$ and $-\tfrac{\pi}{3}$) are exactly these two residues, re-weighted. The pole $z=0$ is not a feature of the physics — it is born from the measure $dt=dz/(iz)$. The pole $z=-\tfrac12$ is the physical pole of $1/(5+4\cos t)$. The zero is the statement that these two unrelated-looking poles carry equal and opposite residues.

Open threads ▸

For which numerators $N(z)$ does $\oint \frac{N(z)}{z(2z+1)(z+2)}\,dz$ vanish? The condition is $\Res_{0}+\Res_{-1/2}=0$, a single linear equation in the coefficients of $N$. Map the kernel.
What does the outside pole $z=-2$ know? Compute $\Res_{z=-2}=+\tfrac12$ and the residue at infinity, and revisit in Rung 4.
Real vs complex residue. Both interior residues here are real; the Interlude below reads a real residue as a source/sink and a complex one as a vortex of the Pólya vector field.

Rung 3 — Number theory: a cyclotomic numerator over a palindromic denominator

Settled core. Two exact identities on the circle.

Numerator. $z^2+z+1=\Phi_3(z)$, the third cyclotomic polynomial (Ireland & Rosen, 1990), whose roots are the primitive cube roots of unity $e^{\pm 2\pi i/3}$. On $\lvert z\rvert=1$,

\[1+2\cos t=e^{-it}+1+e^{it}=\sum_{k=-1}^{1}e^{ikt}=D_1(t),\]

the order-1 Dirichlet kernel. The numerator is not arbitrary; it is the simplest nonconstant Dirichlet kernel.

Denominator. $5+4\cos t=\lvert 2+e^{it}\rvert^2$ — check $(2+\cos t)^2+\sin^2 t=5+4\cos t$. As a polynomial, $2z^2+5z+2$ has coefficient vector $(2,5,2)$, which is palindromic; palindromic polynomials have reciprocal root sets, here ${-\tfrac12,-2}$ with $(-\tfrac12)(-2)=1$. One root sits inside the unit circle, its reciprocal sits outside — a mirror pair across $\lvert z\rvert=1$. Exactly one physical pole is therefore enclosed, which is what makes a single nonzero residue available to cancel the measure’s residue at $0$.

Proposition (Fourier reading, best paper-ready). Expand the denominator’s reciprocal as a Fourier/Laurent series (Stein & Shakarchi, 2003), $\frac{1}{5+4\cos t}=\sum_{n\in\Z}c_n e^{int},\qquad c_n=\frac13\left(-\frac12\right)^{\lvert n\rvert}.$ Because $1+2\cos t=D_1(t)$ selects the modes $n\in{-1,0,1}$, $\frac{1}{2\pi}\int_0^{2\pi}\frac{1+2\cos t}{5+4\cos t}\,dt=c_0+c_1+c_{-1}=c_0+2c_1=\frac13-\frac26=0.$

This is the cleanest why: the integral is one weighted sum of Fourier coefficients of a Poisson-type kernel, and that sum is zero because $c_0=\tfrac13$ and $c_{\pm1}=-\tfrac16$.

Open threads ▸

Generalize the kernel. Replace $\Phi_3$ by $\Phi_n$, or $\lvert 2+e^{it}\rvert^2$ by $\lvert m+e^{it}\rvert^2$ (Fourier coefficients $c_n=\frac{1}{m^2-1}(-1/m)^{\lvert n\rvert}$ up to scale). For which $(\text{numerator degree},m)$ does the weighted coefficient sum still vanish?
Poisson kernel link. $1/(5+4\cos t)$ is, up to scale, the Poisson kernel at radius $r=-\tfrac12$ (Stein & Shakarchi, 2003). State the integral as “the Dirichlet kernel tests the Poisson kernel,” and connect to summability of Fourier series.

Rung 4 — Topology: winding numbers and a de Rham class

Settled core. The residue theorem is really a pairing between a cycle (the contour, an element of homology) and a cocycle (the meromorphic 1-form, an element of de Rham cohomology). Write the contour’s winding number about each pole, $n(\gamma,p)=\frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-p}$. For $\gamma={\lvert z\rvert=1}$:

\[n(\gamma,0)=1,\quad n(\gamma,-\tfrac12)=1,\quad n(\gamma,-2)=0,\]

so the general residue formula $\int=\sum_p n(\gamma,p)\,\Res_p$ reads, numerically,

\[\int=1\cdot\tfrac12+1\cdot(-\tfrac12)+0\cdot\tfrac12=0.\]

The form $\omega=\dfrac{z^2+z+1}{z(2z+1)(z+2)}\,dz$ lives on the thrice-punctured plane $X=\C\setminus{0,-\tfrac12,-2}$, whose first de Rham cohomology is $H^1{\mathrm{dR}}(X)\cong\R^3$ (Bott & Tu, 1982), generated by the three small loops; the class of $\omega$ is its residue vector $(\tfrac12,-\tfrac12,\tfrac12)$ (up to $2\pi i$). In homology, the unit circle is the sum $[\,\lvert z\rvert=1\,]=[\text{loop}_0]+[\text{loop}{-1/2}]$ in $H_1(X)$ (Hatcher, 2002), and the pairing reads off $\tfrac12+(-\tfrac12)=0$.

A consistency check from the sphere: on $\widehat{\C}$ the residues at all poles, including infinity, must sum to zero,

\[\Res_0+\Res_{-1/2}+\Res_{-2}+\Res_{\infty}=\tfrac12-\tfrac12+\tfrac12-\tfrac12=0.\;\checkmark\]

Open threads ▸

Change the cycle. If $\gamma$ winds twice, or encloses only $z=-2$, or encloses nothing, the answer becomes $0$, $\tfrac{2\pi}{?}$, or $0$ respectively — derive each from the homology class, not by recomputing integrals.
Why three punctures give $\R^3$. Sketch the Mayer–Vietoris / generators-and-relations argument and ask what the relation “$\sum$ residues $=0$ on $\widehat\C$” corresponds to in $H^1$ of the sphere.

Rung 5 — Green ⟶ Stokes: the residue theorem as a boundary law

Settled core. Green’s theorem in the plane,

\[\oint_{\partial D}(P\,dx+Q\,dy)=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA,\]

applied to $f(z)\,dz=(u+iv)(dx+i\,dy)$ with $f$ holomorphic, has integrand curl $\equiv 0$ by the Cauchy–Riemann equations — that is Cauchy’s theorem $\oint f\,dz=0$. When $f$ has poles, excise small disks $D_\varepsilon(p_k)$ around them; on the punctured region the curl still vanishes, so

\[\oint_{\partial D} f\,dz=\sum_k \oint_{\partial D_\varepsilon(p_k)} f\,dz=2\pi i\sum_k \Res_{p_k}f.\]

Our integral is precisely the left side with $\partial D={\lvert z\rvert=1}$ deformed onto the two tiny circles around $z=0$ and $z=-\tfrac12$, recovering $2\pi i(\tfrac12-\tfrac12)=0$.

Statement (the generalization, best paper-ready). All three of Green’s theorem, Cauchy’s theorem, and the residue theorem are the single law $\int_{\partial\Omega}\omega=\int_{\Omega}d\omega$ (Stokes’ theorem (Spivak, 1965)(Lee, 2012)) for a 1-form $\omega$ on an oriented manifold-with-boundary $\Omega$. The residue theorem is the case $\Omega=$ disk-minus-punctures, $\omega=f\,dz$, $d\omega=0$ away from the poles.

The two substitutions used earlier are this same machinery in coordinates: $u=\tan(t/2)$ and $z=e^{it}$ are diffeomorphisms, and the factors $dt=\frac{2\,du}{1+u^2}$ and $dt=\frac{dz}{iz}$ are their Jacobians — the multivariable change-of-variables rule, which is itself the pullback that makes Stokes coordinate-free.

Open threads ▸

Lift one dimension. What 2-form $d\omega$ on what surface has our circle integral as its boundary value? Make “the zero is a boundary that bounds” literal.
Discrete Stokes. State the summation-by-parts / lattice analogue and check the vanishing survives discretization (relevant to numerically evaluating such integrals without catastrophic cancellation).
The vector-field instance. The Interlude below works this Green’s-theorem reading out explicitly through the Pólya vector field — circulation plus flux of $(u,-v)$.

Interlude — one line integral, two readings: parametric curve vs vector field

The five lenses split along a hidden seam. Rungs 1–2 read the integral as a parametrized curve; Rung 5 reads it as a vector field. Naming that seam is itself a finding — it explains why the two give the same zero.

Parametric reading. $z(t)=e^{it}$ is a curve $\gamma:[0,2\pi]\to\C$, and

\[\oint_\gamma g(z)\,dz=\int_0^{2\pi} g(z(t))\,z'(t)\,dt,\qquad g(z)=\frac{z^2+z+1}{z(2z+1)(z+2)},\]

is exactly the $dt$-integral of Rungs 1–2 (the original integral is $\tfrac1i\oint_\gamma g\,dz$) — a function traced along a path.

Vector-field reading (the Pólya field). Write $g=u+iv$ and form the Pólya vector field $V=(u,-v)$. Then

\[\oint_C g\,dz=\underbrace{\oint_C(u\,dx-v\,dy)}_{\text{circulation of }V}\;+\;i\underbrace{\oint_C(u\,dy+v\,dx)}_{\text{flux of }V}.\]

Where $g$ is holomorphic the Cauchy–Riemann equations make $V$ simultaneously curl-free and divergence-free; each pole is a point source, sink, or vortex. The residues do the bookkeeping:

\[\text{circulation}=-2\pi\,\mathrm{Im}\!\sum\Res,\qquad \text{flux}=2\pi\,\mathrm{Re}\!\sum\Res.\]

Statement (source–sink reading, best paper-ready). Both interior residues are real ($+\tfrac12$ at $z=0$, $-\tfrac12$ at $z=-\tfrac12$), so neither pole is a vortex: $z=0$ is a pure source of the Pólya field (flux $+\pi$) and $z=-\tfrac12$ a pure sink (flux $-\pi$). They are equal and opposite, so the unit circle encloses zero net flux and zero circulation — and the integral is $0$.

(Checked numerically: $\mathrm{Re}\oint=\mathrm{Im}\oint=0$, and the flux through a tiny circle is $+\pi$ around $z=0$ and $-\pi$ around $z=-\tfrac12$.)

This is the bridge. The elementary/parametric lens and the Green–Stokes/field lens are not two proofs but one object seen two ways: a function dragged once around a loop, or an incompressible irrotational flow with a source and an equal sink inside that loop.

Open threads ▸

When does a vortex appear? A residue with nonzero imaginary part contributes circulation, not just flux. Which numerators $N(z)$ over $z(2z+1)(z+2)$ produce a complex interior residue, and what does the resulting swirl look like?
Re-winding. $z=e^{int}$ traverses the circle $n$ times and the answer stays $0$; track separately what $n$ does to circulation vs flux (it scales the winding-number weight of Rung 4).
Area form. Green’s theorem turns each part into a double integral over the disk; is there a manifestly-zero integrand on the punctured disk (away from the source and sink) that exhibits the cancellation without evaluating residues?

B — The same zero, five windows

Reading the ladder bottom-up collapses it; here are the five lenses side by side, each as object → why-zero → the one computation.

Lens	What it sees	Why it is zero	The one line
Calculus	a rational function on $\R$	two arctangents with equal limits	$\big[\arctan u-\arctan\tfrac u3\big]_{-\infty}^{\infty}=\pi-\pi$
Complex analysis	poles inside $\lvert z\rvert=1$	interior residues cancel	$\Res_0+\Res_{-1/2}=\tfrac12-\tfrac12$
Number theory	$\Phi_3$ over a palindrome	a Fourier-coefficient sum cancels	$c_0+2c_1=\tfrac13-\tfrac26$
Topology	a class in $H^1(X)$	the cycle pairs to zero	$1\cdot\tfrac12+1\cdot(-\tfrac12)$
Green–Stokes	$\int_{\partial\Omega}\omega$	$\omega$ is closed off the poles	$2\pi i\sum\Res=0$

A sixth row would be the bridge itself — parametric curve ⇄ Pólya vector field (the Interlude above) — the seam along which the elementary and the Green–Stokes lenses turn out to be one object.

Synthesis (kept deliberately short). The five “whys” are not five coincidences. The Fourier-coefficient identity $c_0+2c_1=0$ (number theory) is the residue cancellation (complex analysis), is the homology pairing evaluating to zero (topology), is a boundary integral of a closed form (Stokes); the calculus computation is the same statement after a rationalizing diffeomorphism. One object, one zero, five faithful descriptions — and each rung’s open threads point to the family of integrals where the zero either persists or breaks.

References

2012

Introduction to Smooth Manifolds

John M. Lee

2012

Stokes’ theorem on manifolds

2003

Fourier Analysis: An Introduction

Elias M. Stein and Rami Shakarchi

2003

Dirichlet and Poisson kernels
Complex Analysis

Elias M. Stein and Rami Shakarchi

2003

Residues and the Poisson kernel

2002

Algebraic Topology

Allen Hatcher

2002

Fundamental group and homology of the punctured plane

1990

A Classical Introduction to Modern Number Theory

Kenneth Ireland and Michael Rosen

1990

Cyclotomic polynomials

1982

Differential Forms in Algebraic Topology

Raoul Bott and Loring W. Tu

1982

de Rham cohomology; residue as a cohomology pairing

1979

Complex Analysis

Lars V. Ahlfors

1979

Residue theorem and the argument principle

1965

Calculus on Manifolds

Michael Spivak

1965

Stokes’ theorem; the unifying boundary law