Question

\[\int\frac{\text d v}{\sqrt{1+v^2}}\]

Answer 1

?

Answer 2

integrate

Answer 3

look in the back of the text. your integrand is the derivative of a famous function

Answer 4

in other words there are no steps here, no substitutions, no parts, no nothing

Answer 5

i dont want \[=\sinh^{-1}(x)+c\]

Answer 6

what would you prefer?

Answer 7

i mean it is going to be that in any case

Answer 8

you could make \(v=\tan(x)\) to get rid of the radical

Answer 9

get
\[\int \frac{sec^2(x)dx}{\sqrt{\tan^2(x)+1}}\]
\[\int \sec(x)dx\] etc

Answer 10

i want to show it
\[=\ln \left(v+\sqrt{1+v^2}\right)\]

Answer 11

not sure why though
ok then this should work

Answer 12

anti derivative of secant is \(\ln(\sec(x)+\tan(x)\) substitute back and you have your answer

Answer 13

will that work out ok?

Answer 14

\[\int\limits\frac{1}{\sqrt{1+v^{2}}} dx\]
have you tried \(1+v^2=y^2\) => \(2xdx=2ydy\)

Answer 15

http://math2.org/math/integrals/more/sec.htm

i assume you 've probably forgotten the integration of secx

Answer 16

i dont remember ever doing that @rs32623

Answer 17

I couldn't get either of those methods to the end
fortunately integral 25 in the back of my book has \[\int\frac{1}{\sqrt{x^2\pm a^2}}=\ln|x+\sqrt{x^2\pm a^2}| +c\]

Answer 18

90% of what you learn in calc 2 is in the back two pages of the book, which is why that is my least favorite subject in all of math

Answer 19

i don't see a point why can't we reach the result by putting the value of secx & tanx in the method described by satelite73 respectively (1+v)^2 & v..

Answer 20

i tried that

Answer 21

we can, but it hinges on knowing that \(\int \sec(x)dx=\ln(\sec(x)+\tan(x))\)

Answer 22

oh we can do that easily

Answer 23

where does that come from

Answer 24

@UnkleRhaukus i've already provided the link for that in my first reply..

Answer 25

sorry @rs32623 i read that page i just didn't understand it very well

Answer 26

if you grant that
\[\int\sec(x)dx=\ln(\sec(x)+\tan(x))\] substitute back. since \(v=\sec(x)\iff x=\sec^{-1}(x)\) we know
\[tan(x)=\tan(\sec^{-1}(x))=\sqrt{1+v^2}\]
substitute back and get
\[\ln(v+\sqrt{1+v^2})\] as required

Answer 27

the tricky part is finding \(\int \sec(x)dx\)

Answer 28

@UnkleRhaukus is the substituting back part ok? that is a set required in all trig substitutions

Answer 29

*step

Answer 30

where did you miss? it was pretty simple..starting from multiplying & dividing by secx+tanx & assuming secx+tanx=u & differentiating both side wrt x & u respectively we have secx(secx+tanx)dx=du so that the last terms is du/u..

Answer 31

as for finding the anti derivative of secant, there is a rather stupid gimmick you can use, namely multiplying top and bottom by \(\sec(x)+tan(x)\) but if you haven't seen it would not know it. there is a more (to my mind) satisfactory method, which is to multiply top and bottom by \(\cos(x)\) replace \(cos^2(x)\) in the denominator by \(1-\sin^2(x)\) and then use partial fractions.
both are rather gimmicky, but that is the nature of finding anti derivative s

Answer 32

that is why, as i said. 90% of what you need in calc 2 is in the back two pages of the book, and frankly now that we have computer algebra systems like maple and mathematica (wolfram) we don't even need those two pages.
a collection of show offy stupid gimmicks, so professors can say
look i can integrate this
look i can integrate that

when the truth of the matter is that if you write down a function the probability it has a nice closed form for an anti derivative is zero

Answer 33

ok, now that i have finished my rant, it is clear how to go from
\[\ln(\sec(x)+\tan(x)\] to
\[\ln(v+\sqrt{1+v^2})\]?

Answer 34

all this confusion and trig identities AGH
i was just trying to get between step in this ODE

Answer 35

it is so trivial, and i am sorry i asked

Answer 36

yeah they are just stating it as if it is known that is all, but as you can see the step is either trivial if you look in the back of the book or
amazingly annoying if you do it by hand

Answer 37

i wrote that by the way

Answer 38

oh yes, integral 25!!

Answer 39

yeah int # [25]

Answer 40

i am glad that there are some people out there that can fill in the details when necessary

Answer 41

thanks