Question

Integrate dx/(1+sqrt(1-x^2))

Answer 1

Do I need to rationalize it?

Answer 2

thats what I was going to write .

Answer 3

multiply it by 1+sqrt(1-x^2)/1+sqrt(1-x^2)?

Answer 4

nop, in rationalizing, you multiply divide by the conjugate
the conjugate here is 1-sqrt(1-x^2)

Answer 5

I thought I was typing "-" lol

Answer 6

So I got integral of 1- sqrt(1-x^2)/x^2 dx

Answer 7

the original problem is integrate dx/1+ sqrt(1-x^2). Can I sue that identity there?

Answer 8

no.

Answer 9

So I really need to use the rationalization

Answer 10

Yeah lol,i read the problem wrong. -_- haha sorry:(

Answer 11

from rationalizing it, I got integral of 1- sqrt(1-x^2)/x^2 dx

Answer 12

oh lol something's wrong i guess

Answer 13

now you can split the integral like this :
integral (1/x^2) - integral (sqrt(1-x^2)/x^2)

first part is easy, for second part, perhaps integration by parts might help ?

Answer 14

I'm now at that. Thanks.

Answer 15

oh wait
for 2nd part, do this
substitute x=sint
this'll make things much easier

Answer 16

\[\large \frac{1}{1+\sqrt{1-x^2}}\cdot \frac{1-\sqrt{1-x^2}}{1-\sqrt{1-x^2}}\]\[\large \frac{1-\sqrt{1-x^2}}{1-(1-x^2)} \]\[\large \frac{1-\sqrt{1-x^2}}{x^2}\]

Answer 17

I've splitted it now like shubhamsrg said

Answer 18

hmm... That's what i got after multiplying by the conjugate.

Answer 19

Ok.

Answer 20

\[\large \int \left( \frac{1-\sqrt{1-x^2}}{x^2}\right)dx \]\[\large \int \frac{1}{x^2}dx - \int \left(\frac{\sqrt{1-x^2}}{x^2}\right)dx \]

Answer 21

What should I do with the second part lol

Answer 22

as I said above, substitute x= sint

Answer 23

didn't see your second reply, sorry

Answer 24

that t is just a variable?

Answer 25

yes

Answer 26

\[\large x= \sin(t) \ , \ dx = \cos(t)dt \]\[\large \int \left(\frac{\sqrt{1-\sin^2(t)}}{\sin^2(t)} \right)dt\]

Answer 27

yay

Answer 28

nop a slight mistake there

Answer 29

uhm, where?

Answer 30

it should be cost dt

Answer 31

the dt

Answer 32

yeah. saw that now

Answer 33

yeah forgot to include that.

Answer 34

sorry -_-

Answer 35

so it will be integral of cot^2 u du

Answer 36

Yes.

Answer 37

yes, thats right
rest should be easier

Answer 38

and \[\large t= \sin^{-1}(t)\]

Answer 39

note that cot^2 u = cosec^2 u -1

and also derivative of cot u is -cosec^2 u

Answer 40

\[\large \int\limits \left(\frac{\sqrt{1-\sin^2(t)}}{\sin^2(t)} \right)(\cos(t))dt\]\[\large \int\limits \left(\frac{\cos^2(t)}{\sin^2(t)} \right)dt\]\[\large \int\limits \left(\cot^2(t) \right)dt\]\[\large -\csc^2(t) +c\]\[\large -\csc^2(\sin^{-1}(t))+c \]

Answer 41

BLEH noooo!!!

Answer 42

.-.

Answer 43

NOOO!!! hold on lol.

Answer 44

I'm confused lol
After I integrate cot^2 u du, how do I substitute it back to x?

Answer 45

\[\large \int\limits \cot^2(t)dt\]\[\large \int\limits(-1 + \csc^2(t))dt\]\[\large -t -\cot(t)+c\]

Answer 46

That's the tricky part, subbing back.

Answer 47

I'll be using arc tan something?

Answer 48

Create a triangle, that's what i would do.

Answer 49

my final answer is -1/x + sin^-1(x) + sqrt(1-x^2)/x + C

Answer 50

GOODJOB!!!!!!!

Answer 51

I'm writing it up using equation editor.

Answer 52

Thank you both. anyway, I can't seem to verify my newly made account here , I've sent the verification link to my email may times

Answer 53

Can I also ask questions in Physics here?

Answer 54

\[\large x= \sin(t) \ \therefore \ t = \sin^{-1}(x)\]\[\large \cot(t) = \frac{\cos(t)}{\sin(t)}=\frac{\sqrt{1-x^2}}{1}\cdot \frac{1}{x}= \frac{\sqrt{1-x^2}}{x} \]\[ -t-\cot(t)+c =-\sin^{-1}(x)-\frac{\sqrt{1-x^2}}{x} +c = -\frac{x\sin^{-1}(x)-\sqrt{1-x^2}}{x}+c \]

Answer 55

I'm lagging hardcore.

Answer 56

And there is a physics section on OS,too. just click "find more subjects"

Answer 57

Okay. Thank you! I'll close this now :)

Answer 58

okie dokie.

Answer 59

shouldn't you just start with the trig substitution right at the outset? surely much easier to do?

Answer 60

Yes you are right, lol. Or use hyperbolic subs....

Answer 61

@Jhannybean hypernice!