Tìm nguyên hàm \(\int {\frac{1}{{\cos x}}dx} \).

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

Đặt t = sin x (–1 < t < 1)

Suy ra: cosxdx = dt

Ta có: \(\int {\frac{{\cos x}}{{1 - {{\sin }^2}x}}} dx = \int {\frac{{dt}}{{1 - {t^2}}} = \int {\frac{{dt}}{{\left( {1 - t} \right)\left( {1 + t} \right)}}} } \)

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

\[ = \frac{1}{2}\left( { - \ln \left( {1 - t} \right) + \ln \left( {1 + t} \right)} \right)\]

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