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b) Ta có
$ \frac{\sin2x}{\sin^{2}x + 2\cos^{2}x}= \frac{\sin2x}{1+\cos^{2}x}= -\frac{(1+\cos^2x)'}{1+\cos^{2}x}$
Suy ra
$\int\limits_{0}^{\pi/4} \frac{\sin2x}{1+\cos^{2}x}dx=\int\limits_{0}^{\pi/4} -\frac{(1+\cos^2x)'}{1+\cos^{2}x}dx=\left[ {-\ln(1+\cos^{2}x)} \right]_{0}^{\pi/4}=\ln\frac{4}{3}$
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