# given $\cosh u = x$ find $\sinh u$

I'm asked to show that:$\newcommand{\arcosh}{\operatorname{arcosh}}$

$\int{x \arcosh x}dx = \frac{1}{4}(2x^2 -1)\arcosh x - \frac{1}{4}x\sqrt{x^2 -1} + C$

If I integrate by parts:

let $u = \arcosh x \Rightarrow \dfrac{du}{dx} = \dfrac{1}{\sqrt{x^2 -1}}$

and let $\dfrac{dv}{dx} = x \Rightarrow v = \frac{1}{2}x^2$

using the formula for integration by parts and rearranging gives:

$I = \frac{1}{2}x^2 \arcosh x - \frac{1}{2}\int{\sqrt{x^2 - 1} + \dfrac{1}{\sqrt{x^2 -1}}}dx$

If I use the substitution:

let $x = \cosh u$

Using this substitution and rearranging gives:

$I = \frac{1}{2}x^2 \arcosh x - \frac{1}{2}\int{\cosh^2 u} du$

$\Rightarrow I = \frac{1}{2}x^2 \arcosh x - \frac{1}{4}[\sinh u \cosh u + u] + C$

I want to find this in terms of $x$. To eliminate $u$, I will find $\sinh u$ in terms of $x$.

I know that $\cosh u = x$

$\Rightarrow \cosh^2 u = x^2$

$\Rightarrow \sinh^2 u + 1 = x^2$

$\Rightarrow \sinh^2 u = x^2 - 1$

$\Rightarrow \sinh u = \pm \sqrt{x^2 - 1}$

I can therefore say:

$I = \frac{1}{2}x^2 \arcosh x - \frac{1}{4}[\pm x\sqrt{x^2 - 1} + \arcosh x] + C$

$\Rightarrow I = \frac{1}{4}(2x^2 - 1)\arcosh x \pm \frac{1}{4}x\sqrt{x^2 - 1} + C$

Have I misunderstood something about $\sinh u$?

Could someone please explain why I must ignore the negative option when square rooting $\sinh^2 u$?

Thank you.

• I would say that the notation $\operatorname{arcosh} x$ is rather unusul, I have seen $\operatorname{argcosh} x$. See en.wikipedia.org/wiki/Inverse_hyperbolic_function May 5 '14 at 12:05
• Ahh, thank you. I haven't seen that notation before. $\arcosh x$ is the notation my examining board uses - but they're probably just being odd. May 5 '14 at 12:15

$\sinh u$ is negative when $u$ is negative and positive otherwise whereas $\cosh u$ is always positive. Therefore, the same is true for $\cosh u\sinh u$. To transfer this property to $x \sqrt{x^2 -1}$, we take the square root to be positive and let $x$ do the job.