Primitive of Hyperbolic Cosecant Function/Inverse Hyperbolic Cotangent of Hyperbolic Cosine Form/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int \csch x \rd x = -2 \map {\coth^{-1} } {e^x} + C$


Proof

Let:

\(\ds u\) \(=\) \(\ds \cosh x\)
\(\ds \leadsto \ \ \) \(\ds u'\) \(=\) \(\ds \sinh x\) Derivative of Hyperbolic Cosine


Then:

\(\ds \int \csch x \rd x\) \(=\) \(\ds \int \frac 1 {\sinh x} \rd x\) Definition of Hyperbolic Cosecant
\(\ds \) \(=\) \(\ds \int \frac {\sinh x} {\sinh^2 x} \rd x\) multiplying top and bottom by $\sinh x$
\(\ds \) \(=\) \(\ds \int \frac {\sinh x} {\cosh^2 x - 1} \rd x\) Difference of Squares of Hyperbolic Cosine and Sine
\(\ds \) \(=\) \(\ds \int \frac {\rd u} {u^2 - 1}\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\coth^{-1} u + C\) Primitive of Reciprocal of $\dfrac 1 {x^2 - a^2}$: $\coth^{-1}$ form
\(\ds \) \(=\) \(\ds -\map {\coth^{-1} } {\cosh x} + C\) Definition of $u$

$\blacksquare$