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

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- Jun 2nd 2006, 09:06 PMchanceyIntegrate 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 PMmalaygoelQuote:

Originally Posted by**chancey**

- Jun 2nd 2006, 09:56 PMchanceyQuote:

Originally Posted by**malaygoel**

EDIT: And it it possible to solve without using trig? - Jun 2nd 2006, 10:18 PMCaptainBlackQuote:

Originally Posted by**chancey**

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 PMmalaygoelQuote:

Originally Posted by**chancey**

- Jun 2nd 2006, 10:44 PMchanceyQuote:

Originally Posted by**malaygoel**

Quote:

Originally Posted by**CaptainBlack**

- Jun 2nd 2006, 11:02 PMCaptainBlackQuote:

Originally Posted by**chancey**

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 PMThePerfectHackerQuote:

Originally Posted by**chancey**

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

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 AMTD!Quote:

Originally Posted by**ThePerfectHacker**

$\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 AMThePerfectHackerQuote:

Originally Posted by**TD!**