Integrate an irrational?

**chancey** · June 2nd 2006, 09:06 PM

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

**malaygoel** · June 2nd 2006, 09:43 PM

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

**chancey** · June 2nd 2006, 09:56 PM

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?

**CaptainBlack** · June 2nd 2006, 10:18 PM

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

**malaygoel** · June 2nd 2006, 10:19 PM

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.

**chancey** · June 2nd 2006, 10:44 PM

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 $\frac{dy}{dx}$ ?

**CaptainBlack** · June 2nd 2006, 11:02 PM

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

**ThePerfectHacker** · June 3rd 2006, 07:16 PM

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,
$\int \sec^2u \frac{du}{dx} dx=\int \sec^2 udu=\tan u$

**TD!** · June 4th 2006, 07:12 AM

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 $

**ThePerfectHacker** · June 4th 2006, 09:44 AM

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.

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