Difference between revisions of "Antiderivative of coth"
From specialfunctionswiki
| (2 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\displaystyle\int \mathrm{coth}(z) \mathrm{d}z=\log(\sinh(z)) + C,$$ | $$\displaystyle\int \mathrm{coth}(z) \mathrm{d}z=\log(\sinh(z)) + C,$$ | ||
for arbitrary constant $C$, where $\mathrm{coth}$ denotes the [[coth|hyperbolic cotangent]], $\log$ denotes the [[logarithm]], and $\sinh$ denotes the [[sinh|hyperbolic sine]]. | for arbitrary constant $C$, where $\mathrm{coth}$ denotes the [[coth|hyperbolic cotangent]], $\log$ denotes the [[logarithm]], and $\sinh$ denotes the [[sinh|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== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[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. █
