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