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If I have an integral: $$\int_{0}^{2\pi} \frac{1}{3+2cos(t)^2} = \frac{a\pi}{b}$$

How can I find a and b? What formula do I use?

user975
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  • Two months and 11 questions after, time to learn, imo, the easy directions to properly write mathematics in this site: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-qu%E2%80%8C%E2%80%8Bick-reference – Timbuc Apr 21 '15 at 21:09
  • Thanks for the link. Yes, I will try and learn how to write it up properly. – user975 Apr 21 '15 at 21:19
  • Integrate in the complex plane by writing $\cos{t}=\frac{1}{2}(e^{i t}+e^{-i t})$. By putting $z=e^{i t}$ and noting that $\mathrm d z = i z \mathrm dt$, your integral is transformed into a line integral over the unit circle $|z|=1$, and you may apply the residue theorem! – krvolok Apr 21 '15 at 21:57

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Suppose we are looking to evaluate $$\int_0^{2\pi} \frac{1}{(3+2\cos x)^2} dx.$$

Put $z=e^{ix}$ so that $dz = i e^{ix} \; dx = iz \; dx$ and use $$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$ to get $$\int_{|z|=1} \frac{1}{(3+z+1/z)^2} \frac{1}{iz} dz = \int_{|z|=1} \frac{z^2}{(z^2+3z+1)^2} \frac{1}{iz} dz \\ = \frac{1}{i} \int_{|z|=1} \frac{z}{(z^2+3z+1)^2} dz.$$

The poles are at $$\rho_{1,2} = -\frac{3}{2}\pm\frac{\sqrt{5}}{2},$$ but only $\rho_1$ is inside the contour.

Write $$\frac{1}{i} \int_{|z|=1} \frac{z-\rho_1+\rho_1}{(z^2+3z+1)^2} dz.$$

This has two components, the first is $$\frac{1}{i} \int_{|z|=1} \frac{1}{(z-\rho_1)(z-\rho_2)^2} dz.$$

We get for this component $$2\pi i\times \frac{1}{i} \times \frac{1}{\sqrt{5}^2} = \frac{2}{5} \pi.$$

The second component is $$\frac{\rho_1}{i} \int_{|z|=1} \frac{1}{(z-\rho_1)^2(z-\rho_2)^2} dz \\ = \frac{\rho_1}{i} \int_{|z|=1} \frac{1}{(z-\rho_1)^2(z-\rho_1+\rho_1-\rho_2)^2} dz \\ = \frac{\rho_1}{i} \int_{|z|=1} \frac{1}{(z-\rho_1)^2(z-\rho_1+\sqrt{5})^2} dz \\ = \frac{\rho_1}{5i} \int_{|z|=1} \frac{1}{(z-\rho_1)^2((z-\rho_1)/\sqrt{5}+1)^2} dz \\ = \frac{\rho_1}{5i} \int_{|z|=1} \frac{1}{(z-\rho_1)^2} \sum_{q\ge 0} (-1)^q (q+1) \frac{(z-\rho_1)^q}{\sqrt{5}^q} \; dz \\ = 2\pi i \times \frac{\rho_1}{5i}\times -\frac{2}{\sqrt{5}} = -\frac{4\pi}{5\sqrt{5}} \rho_1 = \frac{6\pi}{5\sqrt{5}} - \frac{2\pi}{5}.$$

Adding it up (i.e. the two contributions) we obtain $$\frac{6\pi}{5\sqrt{5}} = \frac{6\pi\sqrt{5}}{25} .$$

This MSE link has a similar computation.

Addendum. I just realized (consult question edit history) I misread the question (so you get a bonus integral). In fact we seek to evaluate $$\int_0^{2\pi} \frac{1}{3+2\cos^2 x} dx.$$

Put $z=e^{ix}$ so that $dz = i e^{ix} \; dx = iz \; dx$ and use $$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$ to get $$\int_{|z|=1} \frac{1}{3+(z+1/z)^2/2} \frac{1}{iz} dz \\ = \int_{|z|=1} \frac{2}{6+(z+1/z)^2} \frac{1}{iz} dz \\ = \int_{|z|=1} \frac{2}{z^2+8+1/z^2} \frac{1}{iz} dz = \int_{|z|=1} \frac{2z^2}{z^4+8z^2+1} \frac{1}{iz} dz \\ = \frac{2}{i} \int_{|z|=1} \frac{z}{z^4+8z^2+1} dz.$$

The poles here are at $$\rho_{1,2} = \pm \sqrt{-4+\sqrt{15}}$$ and $$\rho_{3,4} = \pm \sqrt{-4-\sqrt{15}}.$$

However only $\rho_{1,2}$ are inside the contour. These poles are simple and the residues are

$$\frac{\rho_{1,2}}{4\rho_{1,2}^3+16\rho_{1,2}} = \frac{1}{4\rho_{1,2}^2+16} = \frac{1}{-16+4\sqrt{15}+16} = \frac{1}{4\sqrt{15}}.$$

The end result is $$2\pi i \times \frac{2}{i} \times \frac{2}{4\sqrt{15}} = \frac{2\pi}{\sqrt{15}} = \frac{2\pi\sqrt{15}}{15} .$$

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