Đặt: $t=\dfrac{\pi}{2}-x \Rightarrow dt=-dx$
Đổi cận: $x=0 \Rightarrow t=\dfrac{\pi}{2}$
$x=\dfrac{\pi}{2} \Rightarrow t=0$
Ta có:
$I=\int\limits_0^{\frac{\pi}{2}}\ln\dfrac{1+\sin x}{1+\cos x}dx$
$=-\int\limits_{\frac{\pi}{2}}^0\ln\dfrac{1+\cos t}{1+\sin t}dt$
$=\int\limits_0^{\frac{\pi}{2}}\ln\dfrac{1+\cos t}{1+\sin t}dt$
$=-\int\limits_0^{\frac{\pi}{2}}\ln\dfrac{1+\sin t}{1+\cos t}dt=-I$
$\Rightarrow 2I=0 \Rightarrow I=0$