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a.
Ta có:
$\int\limits_0^{\pi/2}\frac{\cos xdx}{7-5\sin x-\cos^2x}$
$=\int\limits_0^{\pi/2}\frac{d(\sin x)}{6-5\sin x+\sin^2x}$
$=\int\limits_0^1\frac{dt}{t^2-5t+6}$
$=\int\limits_0^1(\frac{1}{2-t}-\frac1{3-t})dt$
$=\ln\frac{3-t}{2-t}\left|\begin{array}{l}1\\0\end{array}\right.=\ln2-\ln\frac{3}{2}=\ln\frac{4}{3}$
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