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$I=\int\limits_{1}^{e}\frac{dx}{x\sqrt{4-\ln^2x}} =\int\limits_{1}^{e}\frac{1}{\sqrt{1-\left ( \frac{\ln x}{2} \right )^2}}.\frac{1}{2}.\frac{dx}{x} =\int\limits_{1}^{e}\frac{1}{\sqrt{1-\left ( \frac{\ln x}{2} \right )^2}}.d\left ( \frac{\ln x}{2} \right ) $
$=\arcsin \left ( \frac{\ln x}{2} \right )\left | \right._1^e=\frac{\pi}{6}$
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