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Ta có
$\dfrac{\cos x \ln(\sin x)}{\sin^{2}x}=-\dfrac{\cos x -\cos x( \ln(\sin x)+1)}{\sin^{2}x}=-\dfrac{\sin x.( \ln(\sin x)+1)' -(\sin x)'( \ln(\sin x)+1)}{\sin^{2}x}$
suy ra
$ \int\limits_{\pi/4}^{\pi/2} \dfrac{\cos x.\ln(\sin x)dx}{\sin^{2}x} =\left[ {-\dfrac{ \ln(\sin x)+1}{\sin x}} \right]_{\pi/4}^{\pi/2}=-1-\dfrac{\ln 2-2}{\sqrt 2}$
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