\int_{{\pi\over 6}}^{{\pi\over 2}}\frac{x+\sin x}{1+\cos x}\hbox{d}x=\\\Bigg (u=x+\sin x,\quad \hbox{d}v=\frac{\hbox{d}x}{1+\cos x}=\frac{\hbox{d}({x\over 2})}{\cos^2 {x\over 2}}\Rightarrow \hbox{d}u= (1+\cos x)\hbox{d}x,\quad v=\newcommand{\tg}{\mathop{\mathrm{tg}}}\tg{x\over 2}\Bigg)\\(x+\sin x)\newcommand{\tg}{\mathop{\mathrm{tg}}}\tg{x\over 2}|_{\pi/6}^{\pi/2}-\int_{\pi\over 6}^{\pi\over 2}\sin x \hbox{d}x=\\\Bigg[(x+\sin x)\newcommand{\tg}{\mathop{\mathrm{tg}}}\tg{x\over 2}+\cos x\Bigg ]_{\pi/6}^{\pi/2}=\\\frac{(\sqrt{3}+1)\pi}{6}