# Math Help - Integrate an irrational?

1. ## Integrate an irrational?

I'm trying to find the function from:

$\int \frac{1}{\sqrt{1-x^2}} dx$

But I don't know how to integrate with that irrational

2. Originally Posted by chancey
I'm trying to find the function from:

$\int \frac{1}{\sqrt{1-x^2}} dx$

But I don't know how to integrate with that irrational
substitute x= siny and you will find it easy to integrate

3. Originally Posted by malaygoel
substitute x= siny and you will find it easy to integrate
I'm very new to integration, if I substitute I get $\sec y$, but I don't know what this means? Is that the function?

EDIT: And it it possible to solve without using trig?

4. Originally Posted by chancey
I'm very new to integration, if I substitute I get $\sec y$, but I don't know what this means? Is that the function?

EDIT: And it it possible to solve without using trig?
When you make a substitution $x=g(y)$ in a indefinite integration you
have:

$\int f(x)dx=\int f(g(y)) \frac{dx}{dy}dy$.

You have found $f(g(y))$ for the substitution you have made, but you also
need to do the part involving $\frac{dx}{dy}$.

RonL

5. Originally Posted by chancey
I'm very new to integration, if I substitute I get $\sec y$, but I don't know what this means? Is that the function?

EDIT: And it it possible to solve without using trig?
if you are new to calculus, I would advise you to clearly understand the concept of a function which is foundation of calculus.

6. Originally Posted by malaygoel
if you are new to calculus, I would advise you to clearly understand the concept of a function which is foundation of calculus.
I said I was new to integration.

Originally Posted by CaptainBlack
but you also need to do the part involving $\frac{dx}{dy}$
Do you mean $\frac{dy}{dx}$ ?

7. Originally Posted by chancey
I said I was new to integration.

Do you mean $\frac{dy}{dx}$ ?
1. You are quoting from two different sources, with only one attribution.
This is confusing, don't do it. Post the two comments as diffrenet messages.

2. No I meant what I wrote. Look at what we are doing: we put:

$
x=g(y)
$

so:

$\frac{dy}{dx}=g'(y)$.

In the case of your integral you have:

$x=\sin(y)$,

so:

$\frac{dx}{dy}=\cos(y)$,

do you see why this may be usefull?

RonL

8. Originally Posted by chancey
I'm trying to find the function from:

$\int \frac{1}{\sqrt{1-x^2}} dx$

But I don't know how to integrate with that irrational
First rationalize,
$\int \frac{\sqrt{1-x^2}}{1-x^2}dx$
Let, $u=\sin^{-1}(x)$ then,
$\frac{du}{dx}=\sqrt{1-x^2}$
And, $\sec(u)=\frac{1}{\sqrt{1-x^2}}$
Thus, $\sec^2(u)=\frac{1}{1-x^2}$

Express integrand as,
$\int \frac{1}{1-x^2}\cdot \frac{\sqrt{1-x^2}}{1} dx$ thus,
$\int \sec^2u \frac{du}{dx} dx=\int \sec^2 udu=\tan u$

9. Originally Posted by ThePerfectHacker
Let, $u=\sin^{-1}(x)$ then,
$\frac{du}{dx}=\sqrt{1-x^2}$
This derivative isn't correct.

$
\int {\frac{1}{{\sqrt {1 - x^2 } }}dx} \to x = \sin y \Leftrightarrow dx = \cos ydy \to \int {\frac{{\cos y}}{{\sqrt {1 - \sin ^2 y} }}dy}
$

$
\int {\frac{{\cos y}}{{\sqrt {\cos ^2 y} }}dy} = \int {dy} = y + C \to \int {\frac{1}{{\sqrt {1 - x^2 } }}dx} = \arcsin x + C
$

10. Originally Posted by TD!
This derivative isn't correct.

$
\int {\frac{1}{{\sqrt {1 - x^2 } }}dx} \to x = \sin y \Leftrightarrow dx = \cos ydy \to \int {\frac{{\cos y}}{{\sqrt {1 - \sin ^2 y} }}dy}
$

$
\int {\frac{{\cos y}}{{\sqrt {\cos ^2 y} }}dy} = \int {dy} = y + C \to \int {\frac{1}{{\sqrt {1 - x^2 } }}dx} = \arcsin x + C
$
Sorry, I was very tired when I was writing this.