Đây là tích phân liên kết
Hoặc làm theo cách Đăt $x=\dfrac{\pi}{2}-t$
$I=\int \limits_0^{\frac{\pi}{2}} \dfrac{\cos (\dfrac{\pi}{2}-t)}{\sin (\dfrac{\pi}{2}-t) + \cos (\dfrac{\pi}{2}-t)}dt$
$= \int \limits_0^{\frac{\pi}{2}} \dfrac{\sin t}{\sin t +\cos t}dt = \int \limits_0^{\frac{\pi}{2}} \dfrac{\sin x}{\sin x +\cos x}dx =I$
Ta có $I+I = \int \limits_0^{\frac{\pi}{2}} \dfrac{\cos x}{\sin x +\cos x}dx+\int \limits_0^{\frac{\pi}{2}} \dfrac{\sin x}{\sin x +\cos x}dx = \int \limits_0^{\frac{\pi}{2}} dx = x \bigg |_0^{\frac{\pi}{2}} = \dfrac{\pi}{2} \Rightarrow I =\dfrac{\pi}{4}$