Решение.

Применяем формулы тригонометрии и делаем замену переменных .

[tex]\displaystyle \int\limits_{0}^{\frac{\pi}{6}}\dfrac{dx}{cos(2x)}=\int\limits_{0}^{\frac{\pi}{6}}\dfrac{dx}{sin(\frac{\pi}{2}+2x)}=\int\limits_{0}^{\frac{\pi}{6}}\dfrac{dx}{2\, sin(\frac{\pi}{4}+x)\cdot cos(\frac{\pi}{4}+x)}=\\\\\\=\int\limits_{0}^{\frac{\pi}{6}}\dfrac{dx}{2\cdot \dfrac{sin(\frac{\pi}{4}+x)}{cos(\frac{\pi}{4}+x)}\cdot cos^2(\frac{\pi}{4}+x)}=\int\limits_{0}^{\frac{\pi}{6}}\dfrac{1}{2\cdot tg(\frac{\pi}{4}+x)}\cdot \frac{dx}{cos^2(\frac{\pi}{4}+x)}=[/tex]

[tex]\displaystyle =\Big[\ u=tg(\frac{\pi}{4}+x)\ ,\ du=\frac{dx}{cos^2(\frac{\pi }{4}+x)}\ ,\ u_1=tg\frac{\pi}{4}=1\ ,\ u_2=tg\frac{5\pi}{12}\ \Big]=\\\\\\=\frac{1}{2}\int\limits_1^{tg\frac{5\pi}{12}}\frac{1}{u}\cdot du=\frac{1}{2}\int\limits_1^{tg\frac{5\pi}{12}}\frac{1}{u}\cdot du=\frac{1}{2}\cdot ln|u|\ \Big|_1^{tg\frac{5\pi }{12}}=\frac{1}{2}\cdot ln\Big|\, tg\frac{5\pi}{12}\, \Big|-0=\\\\\\=\frac{1}{2}\cdot ln|\, 2+\sqrt3\, |=\frac{1}{2}\cdot ln(2+\sqrt3)[/tex]