1: đặt : x=-t khi đó ta có : I=π/2∫0sin5x1+cosxdx=0∫π/2−sin5t1+cost(−dt)=π/2∫0−sin5t1+costdt=−I⇔2I=0⇔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$