Đặt \(t=cosx, dt=-sinx.dx\) Với x = 0 thì t = 1. Với \(x=\frac{\pi}{3}\Rightarrow t=\frac{1}{2}\) \(sin^3xdx=(1-cos^2x).sinxdx=-(1-t^2)dt\) \(I=\int_{1}^{\frac{1}{2}}\frac{(1-t^2)}{t^2}dt=\int_{1}^{\frac{1}{2}}\left ( \frac{1}{t^2}-1 \right )dt=\left ( -\frac{1}{t}-t \right )\bigg |_{\frac{1}{2}}^1=\frac{1}{2}\)