đặt $x= \dfrac{\pi}{2}-t \Rightarrow dx=-dt$
$I = -\displaystyle \int_{\frac{\pi}{2}}^0 \dfrac{\sqrt{ \sin (\dfrac{\pi}{2}-t) }}{\sqrt{\cos (\dfrac{\pi}{2}-t)} +\sqrt{\sin (\dfrac{\pi}{2}-t)}}dt = \displaystyle \int_0^{\frac{\pi}{2}} \dfrac{\sqrt{ \cos t}}{\sqrt{\cos t} +\sqrt{\sin t}}dt $
$=\displaystyle \int_0^{\frac{\pi}{2}} \dfrac{\sqrt{ \cos x}}{\sqrt{\cos x} +\sqrt{\sin x}}dx $
$\Rightarrow 2I = \displaystyle \int_0^{\frac{\pi}{2}} \dfrac{\sqrt{ \sin x}}{\sqrt{\cos x} +\sqrt{\sin x}}dx + \displaystyle \int_0^{\frac{\pi}{2}} \dfrac{\sqrt{ \cos x}}{\sqrt{\cos x} +\sqrt{\sin x}}dx = 1$
Vậy $I =\dfrac{1}{2}$