Tính tích phân sau

$A=\int_{0}^{\frac{\pi }{3}}{\frac{\sin x}{\sqrt{1+6\cos x}}}.dx\breve{a}t:t=\sqrt{1+6\cos x}\Rightarrow {{t}^{2}}=1+6\cos x$

$\Rightarrow 2tdt=-6\sin x.dx\Rightarrow \sin x=-\frac{1}{3}tdt;x=0\Rightarrow t=\sqrt{7};x=\frac{\pi }{3}\Rightarrow t=2$

$\Rightarrow A=\int_{\sqrt{7}}^{2}{-\frac{t}{3}.\frac{1}{t}.dt}=\int_{2}^{\sqrt{7}}{\frac{1}{3}.dt}=\left. \frac{1}{3}t \right|_{2}^{\sqrt{7}}=\frac{1}{3}\left( \sqrt{7}-2 \right)$