66We have
I=\int_0^\pi x\sqrt{1+|\cos x|}\ \mathrm dx=\int_0^\pi (\pi-x)\sqrt{1+|\cos x|}\ \mathrm dx
=\pi\int_0^\pi \sqrt{1+|\cos x|}\ \mathrm dx-\int_0^\pi x\sqrt{1+|\cos x|}\ \mathrm dx
=\pi\int_0^\pi \sqrt{1+|\cos x|}\ \mathrm dx-I
Hence
2I=\pi\int_0^\pi \sqrt{1+|\cos x|}\ \mathrm dx
But
\int_0^\pi \sqrt{1+|\cos x|}\ \mathrm dx =\int_0^{\pi/2} \sqrt{1+\cos x}\ \mathrm dx+\int_{\pi/2}^\pi \sqrt{1-\cos x}\ \mathrm dx =2 + 2 =4
Hence I=2\pi
So (C) is the correct one.
