# Integrals and hyperbolic trigonometry

#### tahirimanov19

$$\displaystyle \int \frac{x \cdot \cosh (x)}{(\sinh (x))^2}dx = ?$$
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
Can you post your working in full? I suggest denoting the inverse functions by $\csch^{-1}(u)$, etc.

topsquark

#### tahirimanov19

$$\displaystyle \frac{x \cdot \cosh (x)}{(\sinh(x))^2} = \coth (x) \cdot \csch(x)$$

$$\displaystyle u=\csch(x) \Rightarrow du=-\coth(x) \cdot \csch(x) \cdot dx$$. 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, $$\displaystyle - \int \csch^{-1}(u)du = -u \cdot \csch^{-1}(u) - \int \frac{du}{\sqrt{u^2+1}}=-u \cdot \csch^{-1}(u) - \sinh^{-1} (u)$$.

topsquark

#### skipjack

Forum Staff
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.

topsquark

#### tahirimanov19

I forgot to put it, but the rest is correct. $$\displaystyle x=\csch^{-1} (u)$$.

#### skipjack

Forum Staff
The derivative of $\csch^{-1}(u)$ is $$\displaystyle -\frac{1}{u^2\sqrt{1 + 1/u^2}}$$, so take care when $u$ is negative.

$\sinh^{-1}(u) = \sinh^{-1}(\csch(x))$

Also, $\sinh^{-1}(u) = \ln\left(u + \sqrt{1 + u^2}\right)$ and $1 + \csch^2(x) = \coth^2(x)$.

topsquark