Primitive of Hyperbolic Secant Function/Arctangent of Exponential Form

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Theorem

$\ds \int \sech x \rd x = 2 \map \arctan {e^x} + C$


Proof

Let:

\(\ds \int \sech x \rd x\) \(=\) \(\ds \int \frac 2 {e^x + e^{-x} } \rd x\) Definition 1 of Hyperbolic Secant
\(\ds \) \(=\) \(\ds \int \frac {2 e^x} {e^{2 x} + 1} \rd x\) multiplying top and bottom by $e^x$

Let:

\(\ds u\) \(=\) \(\ds e^x\)
\(\ds \leadsto \ \ \) \(\ds u'\) \(=\) \(\ds e^x\) Derivative of Exponential Function


Then:

\(\ds \int \sech x \rd x\) \(=\) \(\ds \int \frac {2 \rd u} {u^2 + 1}\) Integration by Substitution
\(\ds \) \(=\) \(\ds 2 \arctan u + C\) Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form
\(\ds \) \(=\) \(\ds 2 \map \arctan {e^x} + C\) Definition of $u$

$\blacksquare$


Also see


Sources

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