Difference between revisions of "Antiderivative of coth"

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==Theorem==
<strong>[[Antiderivative of coth|Theorem]]:</strong> The following formula holds:
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The following formula holds:
$$\displaystyle\int \mathrm{coth}(z)dz=\log(\sinh(z)),$$
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$$\displaystyle\int \mathrm{coth}(z) \mathrm{d}z=\log(\sinh(z)) + C,$$
where $\mathrm{coth}$ denotes the [[coth|hyperbolic cotangent]], $\log$ denotes the [[logarithm]], and $\sinh$ denotes the [[sinh|hyperbolic sine]].
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for arbitrary constant $C$, where $\mathrm{coth}$ denotes the [[coth|hyperbolic cotangent]], $\log$ denotes the [[logarithm]], and $\sinh$ denotes the [[sinh|hyperbolic sine]].
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<strong>Proof:</strong> █  
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==Proof==
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By definition,
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$$\mathrm{coth}(z) = \dfrac{\mathrm{cosh}(z)}{\mathrm{sinh}(z)}.$$
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Let $u=\mathrm{sinh}(z)$ and use the [[derivative of sinh]], [[u-substitution]], and the definition of the [[logarithm]] to derive
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$$\begin{array}{ll}
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\displaystyle\int \mathrm{coth}(z) \mathrm{d}z &= \displaystyle\int \dfrac{1}{u} \mathrm{d} u \\
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&= \log \left( \mathrm{sinh}(z) \right) + C,
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\end{array}$$
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as was to be shown.
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==References==
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[[Category:Theorem]]
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[[Category:Proven]]

Latest 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