# Integrate an irrational?

• Jun 2nd 2006, 09:06 PM
chancey
Integrate an irrational?
I'm trying to find the function from:

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

But I don't know how to integrate with that irrational
• Jun 2nd 2006, 09:43 PM
malaygoel
Quote:

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

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

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

substitute x= siny and you will find it easy to integrate
• Jun 2nd 2006, 09:56 PM
chancey
Quote:

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 $\displaystyle \sec y$, but I don't know what this means? Is that the function?

EDIT: And it it possible to solve without using trig?
• Jun 2nd 2006, 10:18 PM
CaptainBlack
Quote:

Originally Posted by chancey
I'm very new to integration, if I substitute I get $\displaystyle \sec y$, 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 $\displaystyle x=g(y)$ in a indefinite integration you
have:

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

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

RonL
• Jun 2nd 2006, 10:19 PM
malaygoel
Quote:

Originally Posted by chancey
I'm very new to integration, if I substitute I get $\displaystyle \sec y$, 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.
• Jun 2nd 2006, 10:44 PM
chancey
Quote:

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.

Quote:

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

Do you mean $\displaystyle \frac{dy}{dx}$ ?
• Jun 2nd 2006, 11:02 PM
CaptainBlack
Quote:

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

Do you mean $\displaystyle \frac{dy}{dx}$ ?

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:

$\displaystyle x=g(y)$

so:

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

In the case of your integral you have:

$\displaystyle x=\sin(y)$,

so:

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

do you see why this may be usefull?

RonL
• Jun 3rd 2006, 07:16 PM
ThePerfectHacker
Quote:

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

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

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

First rationalize,
$\displaystyle \int \frac{\sqrt{1-x^2}}{1-x^2}dx$
Let, $\displaystyle u=\sin^{-1}(x)$ then,
$\displaystyle \frac{du}{dx}=\sqrt{1-x^2}$
And, $\displaystyle \sec(u)=\frac{1}{\sqrt{1-x^2}}$
Thus, $\displaystyle \sec^2(u)=\frac{1}{1-x^2}$

Express integrand as,
$\displaystyle \int \frac{1}{1-x^2}\cdot \frac{\sqrt{1-x^2}}{1} dx$ thus,
$\displaystyle \int \sec^2u \frac{du}{dx} dx=\int \sec^2 udu=\tan u$
• Jun 4th 2006, 07:12 AM
TD!
Quote:

Originally Posted by ThePerfectHacker
Let, $\displaystyle u=\sin^{-1}(x)$ then,
$\displaystyle \frac{du}{dx}=\sqrt{1-x^2}$

This derivative isn't correct.

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

$\displaystyle \int {\frac{{\cos y}}{{\sqrt {\cos ^2 y} }}dy} = \int {dy} = y + C \to \int {\frac{1}{{\sqrt {1 - x^2 } }}dx} = \arcsin x + C$
• Jun 4th 2006, 09:44 AM
ThePerfectHacker
Quote:

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

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

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

Sorry, I was very tired when I was writing this.