A Mini Chain

We have $\begin{aligned} A & = \; \int_0^\pi \ln(1 - 2e\cos x + e^2)\,dx \; = \; \int_0^\pi \ln|e-e^{ix}|^2\,dx \; = \; 2\int_0^\pi \ln|e - e^{ix}|\,dx \; = \; \int_0^{2\pi}\ln|e-e^{ix}|\,dx \\ & = \; \mathfrak{Re}\left(\int_0^{2\pi} \log(e - e^{ix})\,dx\right) \; = \; \mathfrak{Re}\left(\int_{|z|=1} \log(e-z)\frac{dz}{iz}\right) \; = \; \mathfrak{Re}\left(2\pi\mathrm{Res}_{z=0}\frac{\log(e-z)}{z}\right) \; = \; 2\pi \end{aligned}$ and hence $\begin{aligned} B & = \; \int_0^{2\pi}e^{\cos t} \cos(\sin t)\,dt \; = \; \mathfrak{Re}\left(\int_0^{2\pi}e^{\cos t + i \sin t}\,dt\right) \; = \; \mathfrak{Re}\left(\int_{|z|=1} e^z \frac{dz}{iz}\right) \\ & = \; \mathfrak{Re}\left(2\pi\mathrm{Res}_{z=0}\frac{e^z}{z}\right) \; = \; 2\pi \end{aligned}$ so that $AB = \boxed{4\pi^2}$ .