I = \int\limits_{0}^{\Pi /2} (\cos (\ln (1+\cos x))dx
Đặt \begin{ u = \ln (1+\cos x)}x= \\ y= \end{dv = \cos x} \Rightarrow \begin{du = \frac{\mathrm{-\sin x} }{\mathrm{1+\cos x} x}}x= \\ y= \end{v = \sin x}
\Rightarrow I = \sin x.\ln \left ( 1+\cosx )\left| {} \right|