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$I=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{xdx}{\sin^2x}=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left (\cot x -\cot x+ \frac{x}{\sin^2x} \right )dx$
$= \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cot xdx- \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left (\cot x- \frac{x}{\sin^2x} \right )dx$
$= \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{d(\sin x)}{\sin x}- \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left (x\cot x \right )'dx$
$=\left[ {\ln |\sin x|-x\cot x} \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} $
$=\boxed{\displaystyle{\frac{9-4\sqrt 3}{36}\pi+\frac{1}{2}\ln\frac{3}{2}}}$
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