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$\int\limits_{0}^{\frac{\pi}{3}}\frac{sin^{2}x}{sinx+\sqrt{3}cosx}dx=\frac{1}{2}\int\limits_{0}^{\frac{\pi}{3}}\frac{sin^{2}x}{sin(x+\frac{\pi}{3})}dx$
Đặt:
$x+\frac{\pi}{3}=t \Rightarrow dx=dt$.
$\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}}\frac{sin^{2}(t-\frac{\pi}{3})}{sint}dt=\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}}\frac{\frac{1}{4}sin^{2}t-\frac{\sqrt{3}}{2}sintcost+\frac{3}{4}cos^{2}t}{sint}dt=\frac{1}{4}\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}}sintdt-\frac{\sqrt{3}}{2}\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}}sintdt+\frac{3}{4}\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}}\frac{1}{sint}dt-\frac{3}{4}\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}}sintdt$
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