giup` em bài nay voi

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_iniesta_

[tex]\int\limits_{\pi/6 }^{\pi/2} \frac{sin^2x}{sin3x}dx[/tex]
[TEX][tex]\int\limits_{\pi/6}^{\pi/2} \frac{sin^2x}{ 3sinx -4sin^3x}dx[/tex] [/TEX]
[TEX] \frac{sinx }{3-4sin^2x} = \frac{sinx}{4cos^2x-1}[/TEX]
đặt t =cosx
 
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