Đặt $x=\dfrac{\pi}{2}-t \Rightarrow dx= -dt$
$I=\int_0^{\frac{\pi}{2}} \ln \dfrac{1+\cos t}{1+\sin t} dt=\int_0^{\frac{\pi}{2}} \ln \dfrac{1+\cos x}{1+\sin x} dx$
$\Rightarrow2I= \int_0^{\frac{\pi}{2}} \ln \dfrac{1+\sin x}{1+\cos x} dx+\int_0^{\frac{\pi}{2}} \ln \dfrac{1+\cos x}{1+\sin x} dx$
$=\int_0^{\frac{\pi}{2}} \bigg ( \ln \dfrac{1+\sin x}{1+\cos x} + \ln \dfrac{1+\cos x}{1+\sin x} \bigg)dx=\int_0^{\frac{\pi}{2}} \ln 1 dx = 0 \Rightarrow I=0$