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Ta có:
$ \int\limits_{0}^{\frac{\pi}{4}}\frac{1 +\sin2x}{\cos^{2}x}dx $
$=\int\limits_{0}^{\frac{\pi}{4}}\frac{1}{\cos^{2}x}dx+\int\limits_{0}^{\frac{\pi}{4}}\frac{\sin2x}{\cos^{2}x}dx $
$=\tan x\left|\begin{array}{l}\frac{\pi}{4}\\0\end{array}\right.-\int\limits_{0}^{\frac{\pi}{4}}\frac{d(\cos^2x)}{\cos^{2}x}$
$=1-\ln(\cos^2x)\left|\begin{array}{l}\frac{\pi}{4}\\0\end{array}\right.=1+\ln2$
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