Question

integral of (1+v^2)^-1/2 dv

Answer 1

heard of hyperbolic functions ?

Answer 2

\[\int\limits (1/\sqrt{1+v^2}) dv\]

Answer 3

yeah, got that.
put v= sinh x

Answer 4

dv=?

Answer 5

-cosh x +c

Answer 6

for dv ?
for dv u don't need +c
and dv = cosh x dx

Answer 7

and 1+ sinh^2 x =?

Answer 8

cosh^2 x

Answer 9

dv/sqrt(1+v^2) = cosh x dx /sqrt(1+ sinh^2 x) =cosh x dx/sqrt(cosh^2 x)
= cosh x dx/cosh x = dx
so what is integral of dx ?

Answer 10

got that ^ ?

Answer 11

still chewing

Answer 12

which step u have doubt with ?

Answer 13

Integral of dx is just x + c

Answer 14

yes.
so we had sinh x =v
what is x from here ?

Answer 15

x = sinh^-1 v

Answer 16

so your integral equals
sinh^-1 v+c
got it ?

Answer 17

Ok. I am looking in a text book and their answer looks different. It actually stems from a larger problem. Here is the whole thing:
Solve the differential equation:
\[dv/\sqrt{1+v^2}=dx/x\]

Answer 18

This is after rearranging to separate variables...

Answer 19

and u know how to integrate dx/x right ?

Answer 20

yes

Answer 21

ln(x) + c

Answer 22

so u have
sinh^-1 v = ln x+c

Answer 23

But their depiction of "integrating both sides" looks like this:\[\ln(v+\sqrt{1+v^2}) = \ln(x) + \ln(c)\]

Answer 24

yes, sinh^-1 v can also be written like that, ln(v+sqrt(1+v^2))

Answer 25

both the forms are identical

Answer 26

Ok, so that is a trig property that I should know (or at least can look up)?

Answer 27

yes, u can derive that, if u want
using the definition of sinh x

Answer 28

Ok, got it. thank you for your help!

Answer 29

welcome ^_^