Primitive of Hyperbolic Secant Function/Arctangent of Hyperbolic Sine Form

Theorem

$\ds \int \sech x \rd x = \map \arctan {\sinh x} + C$

We have that:

$\ds \int \sech x \rd x$

$=$

$\ds \int \frac {\d x} {\cosh x}$

Definition 2 of Hyperbolic Secant

$\ds $

$=$

$\ds \int \frac {\cosh x \rd x} {\cosh^2 x}$

multiplying top and bottom by $\cosh x$

$\ds $

$=$

$\ds \int \frac {\cosh x \rd x} {1 + \sinh^2 x}$

Let:

$\ds u$

$=$

$\ds \sinh x$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \cosh x$

Then:

$\ds \int \sech x \rd x$

$=$

$\ds \int \frac {\d u} {u^2 + 1}$

$\ds $

$=$

$\ds \arctan u + C$

$\ds $

$=$

$\ds \map \arctan {\sinh x} + C$

substituting for $u$

$\blacksquare$