Rút gọn như thế này$\frac{(1-cos2x)^2}{sin^22x}=\frac{(2sin^2x)^2}{4sin^2x.cos^2x}=\frac{sin^2x}{cos^2x}=\frac{1}{cos^2x}-1$
Vậy
$\int \frac{(1-cos2x)^2}{sin^22x}dx=tanx-x$
$\Rightarrow \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{(1-cos2x)^2}{sin^22x}dx=(tan\frac{\pi}{3}-\frac{\pi}{3})-(tan\frac{\pi}{4}-\frac{\pi}{4})$
$=\sqrt3-1-\frac{\pi}{12}$