Difference between revisions of "Antiderivative of coth"

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==Proof==
 
==Proof==
 
+
By definition,
 +
$$\mathrm{coth}(z) = \dfrac{\mathrm{cosh}(z)}{\mathrm{sinh}(z)}.$$
 +
Let $u=\mathrm{sinh}(z)$ and use the [[derivative of sinh]], [[u-substitution]], and the definition of the [[logarithm]] to derive
 +
$$\begin{array}{ll}
 +
\displaystyle\int \mathrm{coth}(z) \mathrm{d}z &= \displaystyle\int \dfrac{1}{u} \mathrm{d} u \\
 +
&= \log \left( \mathrm{sinh}(z) \right) + C,
 +
\end{array}$$
 +
as was to be shown. █
 
==References==
 
==References==
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Revision as of 22:54, 24 June 2016

Theorem

The following formula holds: $$\displaystyle\int \mathrm{coth}(z) \mathrm{d}z=\log(\sinh(z)) + C,$$ for arbitrary constant $C$, where $\mathrm{coth}$ denotes the hyperbolic cotangent, $\log$ denotes the logarithm, and $\sinh$ denotes the hyperbolic sine.

Proof

By definition, $$\mathrm{coth}(z) = \dfrac{\mathrm{cosh}(z)}{\mathrm{sinh}(z)}.$$ Let $u=\mathrm{sinh}(z)$ and use the derivative of sinh, u-substitution, and the definition of the logarithm to derive $$\begin{array}{ll} \displaystyle\int \mathrm{coth}(z) \mathrm{d}z &= \displaystyle\int \dfrac{1}{u} \mathrm{d} u \\ &= \log \left( \mathrm{sinh}(z) \right) + C, \end{array}$$ as was to be shown. █

References