Integrating $\operatorname{arsech}$ Just found the following exercise from an old book (Bostock & Chandler: Further Pure Mathematics, Functions, Chapter 4, Misc. Ex Q35(b)):
"Prove that $\int_{\frac{4}{5}}^1 \operatorname{arsech}x = 2 \operatorname{arctan} 2 - \frac{\pi}{2} - \frac{4}{5} \log 2$. "
I get the (indefinite) integral to be $x\operatorname{arsech}x + \arcsin x$ (by parts, letting $dv=1$ and $u=\operatorname{arsech} x$), giving the solution $\frac{\pi}{2} - \arcsin \frac{4}{5} - \frac{4}{5}\log 2 $.
It's not a problem to show $2 \operatorname{arctan} 2 - \frac{\pi}{2} = \frac{\pi}{2} - \arcsin \frac{4}{5}$ but I'm curious to see if there is an alternative approach that gives $\arctan$ immediately because I've not found one yet and I wasn't able to find a direct relation between $\arctan$ and $\arcsin$, especially when they have different arguments.
It wouldn't surprise me if it really was just a fiendish extra step added in for this particular case but thought I'd ask in case someone could spot the solution more directly.
Thanks in advance for any help.
Post-Mortem
Many thanks to Alann for spotting what now feels like an obvious step (aren't they always?). Using his solution, I've come up with the following, for those of you interested (let me know if you think there's an error).
For $0<\alpha \leq 1$
$ 2\arctan\left( \sqrt{\frac{1}{\alpha^2} - 1} + \frac{1}{\alpha} \right) = \pi - \arcsin (\alpha)$ ($\arctan \infty = \frac{\pi}{2}$ should cover the $\alpha = 0$ case). Plotting on Wolfram suggests that there's a more interesting relation for negative values but I'm happy with this at the moment.
I'll forgive myself for not spotting this relation sooner.
 A: I'll assume you know that $\text{sech}^{-1}\left(\frac{4}{5}\right)=\ln(2)$.
One way to derive the $\arctan$ is to begin with the substitution $x=\text{sech}(u)$ ($u\geq 0$) and then integrate by parts. This gives
\begin{align*}
\int_{4/5}^{1}\text{sech}^{-1}(x)\text{ }dx &= \int_{\text{sech}^{-1}(4/5)}^{\text{sech}^{-1}(1)}\text{sech}^{-1}(\text{sech }u)\cdot-\text{sech}(u)\tanh(u)\text{ }du\\
&= \int_{0}^{\text{sech}^{-1}(4/5)}u\text{ }\text{sech}(u)\tanh(u)\text{ }du\\
&= \left[-u\text{ }\text{sech}(u)\right]_{0}^{\text{sech}^{-1}(4/5)}-\int_{0}^{\text{sech}^{-1}(4/5)}-\text{sech}(u)\text{ }du\\
&= -\frac{4}{5}\text{sech}^{-1}(4/5)+\int_{0}^{\text{sech}^{-1}(4/5)}\frac{2}{e^u+e^{-u}}du\\
&= -\frac{4}{5}\ln(2)+2\int_{0}^{\ln(2)}\frac{e^u}{(e^u)^2+1}du
\end{align*}
Making the substitution $t=e^u$ gives
\begin{align*}
-\frac{4}{5}\ln(2)+2\int_{0}^{\ln(2)}\frac{e^u}{(e^u)^2+1}du &= -\frac{4}{5}\ln(2)+2\int_{e^0}^{e^{\ln(2)}}\frac{1}{t^2+1}du\\
&= -\frac{4}{5}\ln(2)+2\left(\tan^{-1}(2)-\tan^{-1}(1)\right)\\
&= 2\tan^{-1}(2)-\frac{\pi}{2}-\frac{4}{5}\ln(2)
\end{align*}
which is the desired result.
