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.
$\sinh u$ 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.