Đặt $x=\sin t \Rightarrow dx=\cos t dt$
$I=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{\cos t .\sqrt{1-\sin^2 t}}{\sin^6 t}dt =\int \dfrac{\cos^2 t}{\sin^6 t}dt =\int \cot^2 t .\dfrac{1}{\sin^2 t}. \dfrac{dx}{\sin^2 t}$
$=-\int \cot^2 t (1+\cot^2 t) d(\cot t)=-\int (\cot^4 t +\cot^2 t) d(\cot t)$ quá dễ rồi