Originally Posted by

**jiboom** the question is

show

$\displaystyle

\int_\frac{4}{5}^1 sech^{-1} x =2tan^{-1} 2- \frac{\pi}{2}-\frac{4ln2}{5}

$

i cant get the first term, i end up with $\displaystyle sin^{-1}(\frac{4}{5})$

i proceeded by parts with

$\displaystyle

u=arsech x =ln(\frac{1+\sqrt{1-x^2}}{x}) u'= \frac{-1}{x\sqrt{1-x^2}}

$

v'=1 dv=x

so im left with integrating $\displaystyle \frac{1}{\sqrt{1-x^2}} $

which gives me $\displaystyle sin^{-1} x$

ive put the values 2artan(2) and arcsin(4/5) in calculator and they are not the same so not sure where im going wrong here.

ALSO as im mentioning arsech my books gives the formula for the log form as $\displaystyle _-^+ ln(\frac{1+\sqrt{1-x^2}}{x}) $

how can this be? for the other inverse hyperbolics they pick the positive, why take both for arsech x? how do i work out arsech (3) say? how do i know whether to pick the = or the - value?