Primitive of Hyperbolic Secant Function/Arcsine Form

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Theorem

$\ds \int \sech x \rd x = \map \arcsin {\tanh x} + C$


Proof

Let:

\(\ds \int \sech x \rd x\) \(=\) \(\ds \int \frac {\sech^2 x} {\sech x} \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {\sech^2 x} {\sqrt {1 - \tanh^2 x} } \rd x\) Sum of Squares of Hyperbolic Secant and Tangent

Let:

\(\ds u\) \(=\) \(\ds \tanh x\)
\(\ds \leadsto \ \ \) \(\ds u'\) \(=\) \(\ds \sech^2 x\) Derivative of Hyperbolic Tangent


Then:

\(\ds \int \sech x \rd x\) \(=\) \(\ds \int \frac {\d u} {\sqrt {1 - u^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \map \arcsin u + C\) Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$: Arcsine Form
\(\ds \) \(=\) \(\ds \map \arcsin {\tanh x} + C\) Definition of $u$

$\blacksquare$


Also see


Sources

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