Tính nguyên hàm \(\int {\frac{{dx}}{{\cos x}}} \).

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

Đặt sinx = t (–1 ≤ t ≤ 1)

Suy ra: cosxdx = dt

A = \(\int {\frac{{dt}}{{1 - {t^2}}} = } \int {\frac{{dt}}{{\left( {1 - t} \right)\left( {1 + t} \right)}} = \frac{1}{2}\int {\left( {\frac{1}{{1 - t}} + \frac{1}{{1 + t}}} \right)dt = \frac{1}{2}\left( { - \ln \left| {1 - t} \right| + \ln \left| {1 + t} \right|} \right)} } \)

Vậy \(\int {\frac{{dx}}{{\cos x}}} = \frac{1}{2}\left[ { - \ln \left( {1 - \sin x} \right) + \ln \left( {1 + \sin x} \right)} \right]\).