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

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 $\pi/\sqrt{a^{2} - b^{2}}$.
