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$\int\limits_{\frac{1}{2}}^{1}\sqrt{\frac{1-x}{1+x}}dx$
=$\int\limits_{\frac{1}{2}}^{1}\frac{1-x}{\sqrt{1-x^2}}dx$
Đặt x=sint => dx=cost.dt
x 1/2 1
t $\Pi/6$ $\Pi/2$
= $\int\limits_{\frac{\Pi}{6}}^{\frac{\Pi}{2}}\frac{(1-sint)cost.dt}{\sqrt{1-sin^2t}}$
= $\int\limits_{\frac{\Pi}{6}}^{\frac{\Pi}{2}}(1-sint)dt$
=$(t+cost)|\begin{matrix} \frac{\Pi}{2}\\ \frac{\Pi}{6} \end{matrix}$
=$\frac{\Pi}{3}-\frac{\sqrt{3}}{2}$
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