Tangent half-angle substitution for $\int_0^{2 \pi} \frac{1- \cos x}{3 + \cos x}$

Question

I want to evaluate the following integral using the tangent half-angle substitution $t = \tan (\frac{x}{2})$: $$\int_0^{2 \pi} \frac{1- \cos x}{3 + \cos x} ~dx$$ However, making the substitution gives me $0$ for each of the limits of integration, which is obviously incorrect. I know that one way to solve this problem is to notice that, by symmetry, the equivalent integral 2$\int_0^{\pi} \frac{1- \cos x}{3 + \cos x} ~dx$ allows the subsitution to work. What are ways to make this substitution work without noticing this symmetry? I know that this general question has been asked on here before, but I am specifically interested in how I can make it work with this substitution.

Answer 1

By this substitution we obtain: $$2\int\limits_0^{\pi}\frac{1-\cos{x}}{3+\cos{x}}dx=2\int\limits_0^{+\infty}\frac{1-\frac{1-t^2}{1+t^2}}{3+\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt=4\int\limits_0^{+\infty}\frac{t^2}{(2+t^2)(1+t^2)}dt=$$ $$=4\int\limits_0^{+\infty}\left(\frac{2}{2+t^2}-\frac{1}{1+t^2}\right)dt=2(\sqrt2-1)\pi.$$

Answer 2

To avoid relying on symmetry, note that \begin{align} I=\int_0^{2 \pi} \frac{1- \cos x}{3 + \cos x} ~dx =\int_0^{2 \pi}\bigg(\sqrt2-1- \frac{\sqrt2(3+\cos x)-4}{3 + \cos x} \bigg)dx\\ \end{align} and \begin{align} \int_0^{2 \pi}\frac{3+\cos x-2\sqrt2}{3 + \cos x} dx =2\tan^{-1}\frac{\sin x}{\cos x+3+2\sqrt2}\ \bigg|_0^{2\pi}=0 \end{align} Thus $$I=\int_0^{2 \pi}(\sqrt2-1)dx=2\pi(\sqrt2-1) $$

Answer 3

I'm not sure whether you would consider this exploiting symmetry, but you can make a substitution to translate the domain to a symmetric interval,

$$I = \int_0^{2\pi} \frac{1 - \cos x}{3 + \cos x} \, dx \stackrel{x=y+\pi}= \int_{-\pi}^\pi \frac{1+\cos y}{3-\cos y} \, dy$$

so that $I$ is now amenable to the tangent substitution,

$$I \stackrel{y=2\arctan z}= \int_{-\infty}^\infty \frac{1+\frac{1-z^2}{1+z^2}}{3-\frac{1-z^2}{1+z^2}} \frac{2}{1+z^2} \, dz = \int_{-\infty}^\infty \left(\frac4{2z^2+1} - \frac2{z^2+1}\right) \, dz$$

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If that doesn't fit your criteria, one may also resort to integrating along the unit circle in $\Bbb C$:

$$\begin{align*}I&= \oint_{|z|=1} \frac{1-\frac{z+\frac1z}2}{3+\frac{z+\frac1z}2} \, \frac{dz}{iz} & z=e^{ix} \\ &= i \oint_{|z|=1} \underbrace{\frac{z^2-2z-1}{z \left(z+3-2\sqrt2\right) \left(z+3+2\sqrt2\right)}}_{=:f(z)} \, dz \\ &= -2\pi \left(\underset{z=0}{\operatorname{Res}} f(z) + \underset{z=-3+2\sqrt2}{\operatorname{Res}} f(z)\right) \\ &= -2\pi \left(1-\sqrt2\right) = \boxed{\left(2\sqrt2-2\right)\pi} \end{align*}$$