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Thread: Integrate an irrational?

  1. #1
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    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
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  2. #2
    Super Member malaygoel's Avatar
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    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
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    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?
    Last edited by chancey; Jun 2nd 2006 at 10:01 PM.
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    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
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  5. #5
    Super Member malaygoel's Avatar
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    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.
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  6. #6
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    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 $\displaystyle \frac{dy}{dx}$ ?
    Last edited by CaptainBlack; Jun 2nd 2006 at 11:04 PM. Reason: Added correct attributions to quoted text
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  7. #7
    Grand Panjandrum
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    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
    Last edited by CaptainBlack; Jun 2nd 2006 at 11:06 PM.
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  8. #8
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    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,
    $\displaystyle \int \sec^2u \frac{du}{dx} dx=\int \sec^2 udu=\tan u$
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  9. #9
    TD!
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    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
    $
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  10. #10
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    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.
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