the question is
show
i cant get the first term, i end up with
i proceeded by parts with
v'=1 dv=x
so im left with integrating
which gives me
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
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?
nice,thank-you. Im not sure why my calc gave me different answers All that is left now is to explain the differen signs. Any ideas ? As i put in my OP my book uses + and - ln{...} for arsech, and + and - for its derivative so it looks like they have chosen the other sign. Can you see why ?
first let us find
equal what
we have two values for
if we choose
what we will get
positive sign
if we want to see what is the realtion between and
this is an angle and it is equal to
if it is not clear to you draw a triangle
take inverse sin function for the both sides
different answer why, I think the problem is in the angles since
I said I think
I wish you understand why and see how it comes
having trouble with this. wofram is saying this is not true.
It gives tan^(-1)[4/3]=0.92729521 = sin^(-1) [4/5]
arctan[4/3] - Wolfram|Alpha
and
tan^(-1) [2]= 1.107148715 = sin^(-1)[2/rt(5)]
arctan[2] - Wolfram|Alpha
something is going awry here. Either im entering the expressions wrongly or they are not equal??!