\(I=\int _{0}^{\frac{\pi }{2} }\frac{\sqrt{\cos x} dx}{\sqrt{\sin x} +\sqrt{\cos x} }\)
Đặt \(x=\frac{\pi }{2} -u\Rightarrow dx=-du\)
Đổi cận: \(x=0\Rightarrow u=\frac{\pi }{2} ;x=\frac{\pi }{2} \Rightarrow u=0\)
\(\Rightarrow I=\int _{0}^{\frac{\pi }{2} }\frac{\sqrt{\cos x} dx}{\sqrt{\sin x} +\sqrt{\cos x} } =\int _{0}^{\frac{\pi }{2} }\frac{\sqrt{\sin x} dx}{\sqrt{\sin x} +\sqrt{\cos x} } \)
\(\Rightarrow 2I=\int _{0}^{\frac{\pi }{2} }\frac{\sqrt{\cos x} dx}{\sqrt{\sin x} +\sqrt{\cos x} } +\int _{0}^{\frac{\pi }{2} }\frac{\sqrt{\sin x} dx}{\sqrt{\sin x} +\sqrt{\cos x} } \)
\(\\ {=\int _{0}^{\frac{\pi }{2} }dx=\frac{\pi }{2} \Rightarrow I=\frac{\pi }{4} } \)