Integrals and hyperbolic trigonometry

Mar 2015
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Universe 2.71828i3.14159
\(\displaystyle \int \frac{x \cdot \cosh (x)}{(\sinh (x))^2}dx = ?\)
By doing substitution(?) \(\displaystyle u= \csch (x)\) I got \(\displaystyle -u \cdot \text{arccsch}(u) - \text{arcsinh}(u)\).

Can someone re-simplify \(\displaystyle \text{arcsinh}(u)\) for me?
 
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skipjack

Forum Staff
Dec 2006
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Can you post your working in full? I suggest denoting the inverse functions by $\csch^{-1}(u)$, etc.
 
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Mar 2015
177
67
Universe 2.71828i3.14159
\(\displaystyle \frac{x \cdot \cosh (x)}{(\sinh(x))^2} = \coth (x) \cdot \csch(x)\)

\(\displaystyle u=\csch(x) \Rightarrow du=-\coth(x) \cdot \csch(x) \cdot dx\). Therefore, \(\displaystyle x=\csch^{-1}(u)\).

So, we can rewrite integral as, \(\displaystyle \int - \csch^{-1} (u) du = - \int \csch^{-1}(u)du\).

Using integration by parts, \(\displaystyle - \int \csch^{-1}(u)du = -u \cdot \csch^{-1}(u) - \int \frac{du}{\sqrt{u^2+1}}=-u \cdot \csch^{-1}(u) - \sinh^{-1} (u)\).
 
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skipjack

Forum Staff
Dec 2006
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Your first equation is incorrect, as there's an $x$ in the numerator on the left-hand side, but not on the right-hand side. I've prefixed the various functions with a backslash for you, so that they render properly.
 
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