1: đặt : x=-t khi đó ta có : I=$\int\limits_{0}^{ \pi /2} \frac{sin^5 x}{1+cos x }dx=\int\limits_{\pi/2}^{0}\frac{-sin^5 t}{1+ cos t }(-dt )=\int\limits_{0}^{\pi/2}-\frac{sin^5 t}{1+cos t} dt =-I\Leftrightarrow 2I=0\Leftrightarrow I=0$
1: đặt : x=-t khi đó ta có : I
=$\int\limits_{0}^{ \pi /2} \frac{sin^5 x}{1+cos x }dx=\int\limits_{\pi/2}^{0}\frac{-sin^5 t}{1+ cos t }(-dt )=\int\limits_{0}^{\pi/2}-\frac{sin^5 t}{1+cos t} dt =-I
$$\Leftrightarrow 2I=0\Leftrightarrow I=0$