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a.
Ta có:
$\int\limits_{\pi/6}^{\pi/2} \frac{1}{\sin ^{2}x + 2\sin x\cos x - \cos^{2}x }dx $
$=\int\limits_{\pi/6}^{\pi/2}\frac{\frac{1}{\sin^2x}dx}{1+2\cot x-\cot^2x}$
$=\int\limits_{\pi/6}^{\pi/2}\frac{d(\cot x)}{\cot^2x-2\cot x-1}$
$=\int\limits_\sqrt3^0\frac{dt}{t^2-2t-1}$
$=\frac{1}{2\sqrt2}\int\limits_\sqrt3^0\left(\frac{1}{t-1-\sqrt2}-\frac1{t-1+\sqrt2}\right)dt$
$=\frac{1}{2\sqrt2}\ln\left|\frac{t-1-\sqrt2}{t-1+\sqrt2}\right|\left|\begin{array}{l}0\\\sqrt3\end{array}\right.=\frac{1}{2\sqrt2}(\ln\frac{1+\sqrt2}{\sqrt2-1}-\ln\frac{\sqrt2+1-\sqrt3}{\sqrt3-1+\sqrt2})$
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