Integration + Trigonometry

Opening the squares and using: $\cos^2A+\sin^2A=1$ and $\cos A\cos B+\sin A\sin B=\cos (A-B)$ , given expression simplifies to: $\int_0^\pi (\sqrt{3+4\cos x+2\cos 2x} )dx$ Now using $\cos 2x=2\cos^2 x+1$ , $\int_0^\pi (\sqrt{1+4\cos x+4\cos^2 x}) dx$ $\int_0^\pi |2\cos x+1| dx$ $\int_0^{\frac{2\pi}{3}}(2\cos x+1)dx-\int_{\frac{2\pi}{3}}^{\pi} (2\cos x+1)dx$ $~~=\dfrac{2\pi}{3}+\sqrt3-(\dfrac{\pi}{3}-\sqrt3)$ $\Large =\dfrac{\pi}{3}+2\sqrt3$ $\huge \therefore ~2+3+3=\boxed{\color{#007fff}{8}}$

Note that the given integral is equivalent to

$\int_{0}^{\pi} |e^{ix}+e^{2ix}+e^{3ix}| dx$

That is,

$\int_{0}^{\pi} |e^{-ix}+1+e^{ix}| dx=\int_{0}^{\pi} |1+2\cos{x}| dx$

Evaluating the above integral, keeping in mind the sign of $1+2\cos{x}$ , the answer comes out to be $\frac{\pi}{3}+2\sqrt{3}$ Hence, $2+2+3=\boxed{8}$