Đặt $x=\dfrac{\pi}{2} -t$
$I=\int_0^{\frac{\pi}{2}} \dfrac{\cos t}{(\sin t +\cos t)^2 }dt=\int_0^{\frac{\pi}{2}} \dfrac{\cos x}{(\sin x +\cos x)^2 }dx$
$\Rightarrow 2I = \int_0^{\frac{\pi}{2}} \dfrac{\sin x +\cos x}{(\sin x +\cos x)^2 }dx=\int_0^{\frac{\pi}{2}} \dfrac{1}{\sin x +\cos x }dx$
$=\dfrac{1}{\sqrt 2} \int \dfrac{1}{\sin (x+\dfrac{\pi}{4})}dx$ còn lại là vớ vẩn