Integration of inverse hyperbolic functions

**arze** · September 1st 2010, 10:46 PM

Prove that:
$\displaystyle \int^1_{\frac{4}{5}} arsech xdx=2\arctan 2-\frac{\pi}{2}-\frac{4}{5}\ln 2$

I found $arsech x = \ln{\frac{1+\sqrt{1+x^2}}{x}}$
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!

**Prove It** · September 2nd 2010, 12:52 AM

Use integration by parts with $u = \textrm{arsech}\,{x}$ and $dv = 1$ .

**arze** · September 2nd 2010, 05:56 PM

$u=arsech x \rightarrow \frac{du}{dx}=-\frac{1}{x\sqrt{1-x^2}}$
$\frac{dv}{dx}=1 \rightarrow v=x$

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

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

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