\(I=\int \frac{{\rm d}x}{\cos x} =\int \frac{{\rm cosx}.{\rm d}x}{\cos ^{2} x} = \int \frac{cos{\rm x}.{\rm d}x}{1-\sin ^{2} x}\)
Đặt \(t=\sin x\Leftrightarrow {\rm d}t=\cos x{\rm d}x\)
Khi đó:
\(I=\int \frac{1}{1-t^{2} } dt =\frac{1}{2} \ln \left|\frac{1+t}{1-t} \right|+C=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x} \right|+C\)