# Integrating $\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$

Can someone please help me integrate $$\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$$ the question says, as a hint, use $\cos\theta = \frac{z + z^{-1}}{2}$ with $|z|=1$. I'm not really sure where to start.

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Let $p(x) = (x-x_1) (x-x_2) \cdots (x-x_n).$ An application of the product rule shows this useful identity (for all $x$ not a root of $p$): $$\frac{p'(x) }{p(x) }= \sum_{k=1}^n \frac{1}{x-x_k}.$$

Now since $\displaystyle \cos t = \frac{e^{it} + e^{-it}}{2}$ we have $$1+ 8\cos^2 \left( \frac{k\pi}{n} \right) = 5 + 2a_k + 2a^{-1}_k$$ where $a_k= \exp(2k\pi i/n),$ which are precisely the roots of $x^n-1.$ Thus, we have (after some partial fractions) that

$$\sum_{k=0}^{n-1} \frac{1}{1+8\cos^2(k\pi/n)} = \frac{1}{6} \sum_{k=0}^{n-1} \frac{1}{\frac{-1}{2} - a_k} - \frac{2}{3} \sum_{k=0}^{n-1} \frac{1}{-2-a_k} .$$

Applying the identity we developed lets us evaluate these sums in closed form, and we have

$$\sum_{k=0}^{n-1} \frac{1}{1+8\cos^2(k\pi/n)} = \frac{2n}{3} \frac{2^{n-1} }{2^n + (-1)^{n+1} } + \frac{1}{6} \frac{n (-1/2)^{n-1} }{ (-1/2)^n -1} .$$

Hence $$\int^1_0 \frac{d\theta}{1+8\cos^2(\pi \theta)} = \lim_{n\to\infty} \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{1+8\cos^2(k\pi/n)} = \frac{1}{3}$$

which immediately implies that your integral has value $2\pi/3.$

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There are two things I would try here; both convert the integral into a rational function.

Hint 1: The first is to let $z=e^{i\theta}$, then $2\cos(\theta)=z+\frac1z$ and $\mathrm{d}\theta=\frac{\mathrm{d}z}{iz}$. This requires using contour integration around the unit circle: $$\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta =\oint\frac{1}{1+2\left(z+\frac1z\right)^2}\frac{\mathrm{d}z}{iz}$$

Hint 2: The second is to use $x=\tan(\theta/2)$, then $\cos(\theta)=\frac{1-x^2}{1+x^2}$ and $\mathrm{d}\theta=\frac{2\mathrm{d}x}{1+x^2}$. This requires changing the interval of integration to $[-\pi,\pi]$:

$$\int_{-\pi}^{\pi}\frac{1}{1+8\cos^2\theta}d\theta =\int_{-\infty}^\infty\frac{1}{1+8\left(\frac{1-z^2}{1+z^2}\right)^2}\frac{2\mathrm{d}x}{1+x^2}$$

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An easy way to evaluate:

$I=\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$

$$I=4\int_0^{\frac{\pi}{2}}\frac{1}{8+\frac{1}{\cos^2\theta}}\frac{d\theta}{\cos^2\theta }= 4\int_0^{\frac{\pi}{2}}\frac{1}{9+\tan^2\theta}d\tan\theta= 4\int_0^{\infty}\frac{1}{3^2+t^2}dt= 2\pi/3$$

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A good method. I was going add a third hint to my answer in this direction; however, there was already a similar answer that has since been deleted by its author. I am not quite sure why it was deleted. – robjohn Apr 17 '12 at 13:26

Transform $$\frac{1}{1+8\cos^2\theta}$$ We have $$\cos\theta=\frac{z+z^{-1}}{2}\Rightarrow\cos^2\theta=\frac{1}{4}\left(z^2+z^{-2}+2\right)$$ Now diferential coeficient $$-\sin\theta\;d\theta=\frac{1}{2}\left(dz-z^{-2}\;dz\right)$$ and i know that $$\sin\theta=\frac{z-z^{-1}}{2}$$[...]

Now, you can finish it… (Tip: Introduce this integral terms)

P.D.: Excuse my English.

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