Question

anti derivative of (x+1)/ (sqrt (1-x^2))

Answer 1

Might be helpful to split up the fraction like this first,\[\large\rm \frac{x+1}{\sqrt{1-x^2}}\quad=\frac{x}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x^2}}\]

Answer 2

The first fraction is a simple u-substitution,
while the second one is some trig stuff.

Answer 3

thats where im confused.. on the trig stuff

Answer 4

could you keep walking me through so I can get an idea for the next problem

Answer 5

You're calling it "anti-differentiation",
but are you familiar with the concept of integration at this point?
I just ask because the steps are little bit advanced from the point where you first learn about "anti-differentiation".

Answer 6

a little bit...

Answer 7

I understand integration actually.

Answer 8

So we want to know how to deal with this:\[\large\rm \int\limits \frac{1}{\sqrt{1-x^2}}~dx\]

Answer 9

Just show me what you do and I'll ask you questions about it

Answer 10

k one sec, website is going really really slow right now :)

Answer 11

So recall that you can't apply a square root when you have a difference or sum.
In general \(\large\rm \sqrt{x^2+y^2}\ne x+y\)

Since we're not able to do that...
we would like to like to some how get rid of the subtraction under the root.
We do that by making use of our Pythagorean Identities from Trigonometry.
\(\large\rm 1-\sin^2x=\cos^2x\)
Notice this equation gives us a way to get rid of subtraction.
\(\large\rm 1+\tan^2x=\sec^2x\)
While this equation gives us a way of ridding the pesky addition sign.

Answer 12

So for our problem, we have subtraction...
we would like to match our problem to that first form I mentioned.
So we would like to replace our x with the sine function.
Sine of a new variable*
\[\large\rm 1-(\color{orangered}{x})^2\quad=1-(\color{orangered}{\sin\theta})^2\quad=\cos^2\theta\]See how we can get rid of the subtraction if we do that?

It's not a normal substitution, so it's weird to get used to :)
You're replacing x with the sine of another variable.
\(\large\rm x=\sin\theta\)

Answer 13

Ok makes sense.

Answer 14

we have to use substitution though

Answer 15

Yes, a special type of substitution :)\[\large\rm \int\limits\frac{1}{\sqrt{1-(\color{orangered}{x})^2}}dx\quad=\int\limits\frac{1}{\sqrt{1-(\color{orangered}{\sin\theta})^2}}dx\quad=\int\limits\frac{1}{\sqrt{\cos^2 \theta}}dx\]There is a still a bunch of work to do from this point.
Yes, I can take the cosine out of the root, good.\[\large\rm =\int\limits \frac{1}{\cos \theta}dx\]But, since we're making a substitution,
we also have to replace the differential dx, to something which involves \(\rm d\theta\).

Answer 16

ok but why did we randomply plug in sin theta

Answer 17

It wasn't random :O
It was specifically chosen so we could get rid of the `subtraction` under the root.

Answer 18

See how no subtraction = easier to take a root?

Answer 19

isn't there like a identity for inverse sin or something like that for usub

Answer 20

Well it wouldn't involve usub at all.
it's just an identity that you look at and go... yup, that's are arcsin(x).

I thought you were asking how to do the problem though :) lol
But yes, you can look it up in a table if you prefer.\[\large\rm \int\limits\frac{1}{\sqrt{a^2-x^2}}dx\quad=\arcsin\left(\frac{x}{a}\right)+C\]So for our problem we would simply have,\[\large\rm \int\limits\frac{1}{\sqrt{1^2-x^2}}dx\quad=\arcsin\left(\frac{x}{1}\right)+C\]

Answer 21

that's our* arcsin(x).

Ooo that typo is killing me lol

Answer 22

I have to go. Can you just give me the answer and I"ll look over it later haha

Answer 23

Ok but thers no substitution though.. :(

Answer 24

:o

Answer 25

Wait so Im not following whats the final answer