Question

how to integrate 1/(1+v^2) dv

Answer 1

substitute v by tan x

Answer 2

check your standard table of integrals
,
dont forget to +C

Answer 3

actually tried that and ended up with integral of secx then when trying to integrate secx, it just got a bit messy...

Answer 4

no, dv= sec^2 x dx, at the denominator it will be 1+tan^2 x= sec^2 x , so those two will be canceled out and you will be left with integration of dx, that is x+c, where tanx=v
so x=arctan v

Answer 5

Ahhh... I see... what happens if I add a square root sign of the 1+v^2

Answer 6

then it will be integration over sec x dx

Answer 7

and how would I approach the integration of secx? substitution of arctan?

Answer 8

multiply and divide by secx+tanx, so numerator= sec^2 x +sec x tan x= d(tan x+ sec x), and the denominator is already sec x+ tanx, so the integration will be log |secx + tanx| +c

Answer 9

ahhh.. ok thanks.. sorry my brain's a bit exhausted with trig. Thanks.

Answer 10

you are welcome! :)

Answer 11

\[I=\int\limits \frac{ 1 }{ 1+v^2 }dv\]put v=tan x,
\[dv=\sec ^2x~dx\]
\[I=\int\limits \frac{ \sec ^2 x ~dx }{ 1+\tan ^2x }=\int\limits \frac{ \sec ^2x~dx }{ \sec ^2x }=\int\limits dx=x+c=\tan^{-1} v+c\]