Đáp án:

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

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

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

• Tương tự:  \( \int \frac{1}{\sin x} \, dx \)