$$\begin{align*}&\int\sec{x}dx\\&=\int\frac{1}{\cos{x}}dx\\&=\int\frac{\cos{x}}{\cos^2{x}}dx\\&=\int\frac{d(\sin{x})}{1-\sin^2{x}}\\&=\frac{1}{2}\int\left(\frac{1}{1-\sin{x}}+\frac{1}{1+\sin{x}}\right)d(\sin{x})\\&=\frac{1}{2}\left(-\ln|1-\sin{x}|+\ln|1+\sin{x}|\right)\\&=\boxed{\frac{1}{2}\ln\frac{1+\sin{x}}{1-\sin{x}}+C}\end{align*}$$