Question

General solution

Answer 1

I can help u

Answer 2

Hint:
I can rewrite your equation as follows:
\[\frac{{\sqrt {1 + {y^2}} }}{y}dy = - \frac{{\sqrt {1 + {x^2}} }}{x}dx\]

Answer 3

i tried doing it but i am having a difficulty in integrating it

Answer 4

what did you try?

Answer 5

i tried supposing 1+y as t and 1+x as u and removed sq rt and instead place the power 1/2

Answer 6

you mean U sub
but why did you do t=1+y while we have 1+y^2 under the radical

Answer 7

try \[t=1+y^2 ~\Longrightarrow dt=2ydy\]

Answer 8

sorry forgot to type it

Answer 9

but my instructor told me that im doing it all wrong

Answer 10

i supposed the same thong

Answer 11

hmm then try by parts

Answer 12

hmm i believe that sub works as well, your instructor may be pointing to something else

Answer 13

here have a look

Answer 14

I would actually attempt integration by part first

Answer 15

hmm you the whole thing as t
i suggested a different thing if you paid attention
\[1+y^2=t \Longrightarrow y=\sqrt{1-t}\]
i see why your prof said you got it wrong

Answer 16

im not good at integration :(

Answer 17

dt=2ydy
so the integral of the LHS becomes
\[\int \frac{\sqrt{t}}{2(\sqrt{1-t})^2}dt=\frac{1}{2}\int \frac{\sqrt{t}}{1-t}dt\]

Answer 18

that's what you get after substituting
try it now

Answer 19

but how did we came up with this ? can explain how we had it

Answer 20

well we set t=1+y^2 so y=sqrt{1-t}
we have dt=2ydy so dy=dt/2y
our integrand is sqrt(1+y^2)/y
substituting y
and 1+y^2 we get
sqrt{t}/sqrt{1-t}
now replacing dy by 1/2dt/y which the same as 1/2 dt/sqrt{1-t}
and you get what i did

Answer 21

is this some type of a rule or property we are following here

Answer 22

how about a trig sub that may be a good alternative

Answer 23

there is not rule
just algebra
we set a sub for our 1+y^2 to get a much better integrand that we can easily integrate

Answer 24

try trig sub
make x=tantheta

Answer 25

i guess that would be faster and better

Answer 26

can show me the solution please

Answer 27

i did not do the solution i'm just trying to give some alternatives
you have to work it yourself

Answer 28

it has some work to it!

Answer 29

i literally confused i am not good with this part of integration i was hoping for easier and understandavle approach

Answer 30

eh it is not gonna be that easy
it needs some work

Answer 31

i just tried the trig sub turns out bad
\[\int \csc\alpha\sec^2 \alpha d\alpha \]
i chose \[y=\tan \alpha \]
after this i'm inclined to integration by parts

Answer 32

i am good at partial differentiation and ordinary too but integration is the most difficult one

Answer 33

can you tell me a site where i can find example similar to this question please

Answer 34

if the answer that you want
here
http://www.wolframalpha.com/input/?i=int+%28sqrt%281%2By%5E2%29%2Fy%29dy

Answer 35

you have to do the same to right hand side with the difference of negative sign

Answer 36

looks like the solution will be accomplished by an integration by parts
set u=1/y du=-1/y^2
dv=sqrt{1+y^2} v=int sqrt{1+y^2}
try to work on it more you might get something

Answer 37

ok ill try lets see :) thanks

Answer 38

no matter what you are required to do IBP

Answer 39

no problem

Answer 40

here something this guys did
which the same thing i tried too above
\[\int \csc \alpha \sec^2 \alpha d\alpha \]
https://answers.yahoo.com/question/index?qid=20090211222753AACuIJB

Answer 41

it again needs IBP lol

Answer 42

:)