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
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?Originally Posted by malaygoel
EDIT: And it it possible to solve without using trig?
When you make a substitution $\displaystyle x=g(y)$ in a indefinite integration youOriginally 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
1. You are quoting from two different sources, with only one attribution.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
First rationalize,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,
$\displaystyle \int \sec^2u \frac{du}{dx} dx=\int \sec^2 udu=\tan u$
This derivative isn't correct.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
$