Primitive of Hyperbolic Secant Function/Arctangent of Half Hyperbolic Tangent Form

Theorem

$\ds \int \sech x \rd x = 2 \map \arctan {\tanh \dfrac x 2} + C$

Proof

Let $u = \tanh \dfrac x 2$.

Then:

\(\ds \int \sech x \rd x\)

\(=\)

\(\ds \int \frac {1 - u^2} {1 + u^2} \frac {2 \rd u} {1 - u^2}\)

Hyperbolic Tangent Half-Angle Substitution

\(\ds \)

\(=\)

\(\ds \int \frac {2 \rd u} {1 + u^2}\)

simplifying

\(\ds \)

\(=\)

\(\ds 2 \arctan u + C\)

Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form

\(\ds \)

\(=\)

\(\ds 2 \map \arctan {\tanh \dfrac x 2} + C\)

substituting for $u$

$\blacksquare$

Sources

• 1960: Margaret M. Gow: A Course in Pure Mathematics ... (previous) ... (next): Chapter $10$: Integration: $10.4$. Standard integrals: Other Standard Results: $\text {(xxvii)}$