[TEX]\int\limits_{\pi\3}^{\pi\2}\frac{dx}{sin 2x - 2 sin x}[/TEX]
[TEX]sin 2x - 2 sin x=2sinx( cosx-1)=-4sin \(\(\frac{x}{2}\) cos \(\(\frac{x}{2}\) sin^2 \(\frac{x}{2}\)[/TEX]
[TEX]\I:=-\frac{1}{4}\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{ cos \(\(\frac{x}{2}\)dx}{sin \(\(\frac{x}{2}\) cos^2 \(\frac{x}{2}\) sin^2 \(\frac{x}{2}\)}[/TEX]
[TEX]t={sin \(\(\frac{x}{2}\)\righ dt= \frac{1}{2} cos\(\(\frac{x}{2}\) dx[/TEX]
[TEX]\righ \I:=-\frac{1}{2} \int_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}}\frac{dt}{t^3(1-t^2)}=\frac{1}{2}\(\frac{1}{2 t^2}+ \frac{ln|1-t^2|}{2}-ln|t|\)\|_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}}=DONE!![/TEX]