Answer
$$\int \operatorname{sech} x \, d x=\tan ^{-1}(\sinh x)+C$$
Work Step by Step
Given $$\int \operatorname{sech} x \, d x $$
Let $u=\sinh x,\ \ du=\cosh xdx $. Then
\begin{align*}
\int \operatorname{sech} x \, d x &= \int \frac{1}{\cosh x}dx\\
&= \int \frac{1}{\cosh^2 x}du\\
&=\int \frac{1}{1+\sinh ^{2} u} d u\\
&=\int \frac{1}{1+u^{2}} du\\
&=\tan ^{-1}u+C\\
&=\tan ^{-1}(\sinh x)+C
\end{align*}
