Đáp án:

• \(\int \frac{1}{\cos 2x} \, dx = \int \frac{1}{\cos^2 x - \sin^2 x} \, dx\)

\(=\int \frac{1}{\cos^2 x (1 - \tan^2 x)} \, dx \quad \text{đặt } t = \tan x\)

\(=\int \frac{1}{1 - t^2} \, dt = \int \frac{1}{(1 - t)(1 + t)} \, dt\)

\(=\frac{1}{2} \int \left( \frac{1}{1 - t} + \frac{1}{1 + t} \right) \, dt\)

\(=\frac{1}{2} \ln \left| \frac{t + 1}{1 - t} \right| + C\)