Giúp em câu tích phân

P

passingby

[TEX]\int\limits_{pi\3}^{pi\2}\frac{dx}{sin 2x - 2 sin x}[/TEX]
\Leftrightarrow[TEX]I=\frac{1}{2}\int_{}^{}\frac{dx}{sinx(cosx-1)[/TEX]
\Leftrightarrow[TEX]I=\frac{1}{2}\int_{}^{}\frac{sinxdx}{sin^2x(cosx-1)}[/TEX]
\Leftrightarrow[TEX]I=\frac{1}{2}\int_{}^{}\frac{d(cosx)}{(1-cosx)^2(1+cosx)}[/TEX]
Đồng nhất => okie :D
 
V

vodichhocmai

[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]
 
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