# Integration of inverse hyperbolic functions

• Sep 1st 2010, 10:46 PM
arze
Integration of inverse hyperbolic functions
Prove that:
$\displaystyle \displaystyle \int^1_{\frac{4}{5}} arsech xdx=2\arctan 2-\frac{\pi}{2}-\frac{4}{5}\ln 2$

I found $\displaystyle arsech x = \ln{\frac{1+\sqrt{1+x^2}}{x}}$
But When I tried to integrate it, it got very messy as well as not being the answer. I've got no idea how else to integrate, so any help will be very much appreciated.
Thanks!
• Sep 2nd 2010, 12:52 AM
Prove It
Use integration by parts with $\displaystyle u = \textrm{arsech}\,{x}$ and $\displaystyle dv = 1$.
• Sep 2nd 2010, 05:56 PM
arze
$\displaystyle u=arsech x \rightarrow \frac{du}{dx}=-\frac{1}{x\sqrt{1-x^2}}$
$\displaystyle \frac{dv}{dx}=1 \rightarrow v=x$

Then the integral
$\displaystyle x arsech x -\int -\frac{1}{\sqrt{1-x^2}}$
$\displaystyle =x arsech x + \arcsin x$

Filling in the values, I get:
$\displaystyle \arctan \frac{4}{3}-\frac{\pi}{2}-\frac{4}{5}\ln 2$
Which is correct except for
$\displaystyle \arctan \frac{4}{3}$