Results 1 to 10 of 10

Math Help - Integrate an irrational?

  1. #1
    Junior Member
    Joined
    May 2006
    From
    Sydney, Australia
    Posts
    44

    Integrate an irrational?

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote 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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2006
    From
    Sydney, Australia
    Posts
    44
    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 \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; June 2nd 2006 at 10:01 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote 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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote 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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2006
    From
    Sydney, Australia
    Posts
    44
    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 \frac{dx}{dy}
    Do you mean \frac{dy}{dx} ?
    Last edited by CaptainBlack; June 2nd 2006 at 11:04 PM. Reason: Added correct attributions to quoted text
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote 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:

    <br />
x=g(y)<br />

    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
    Last edited by CaptainBlack; June 2nd 2006 at 11:06 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote 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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    Quote Originally Posted by ThePerfectHacker
    Let, u=\sin^{-1}(x) then,
    \frac{du}{dx}=\sqrt{1-x^2}
    This derivative isn't correct.

    <br />
\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} <br />

    <br />
\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<br />
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by TD!
    This derivative isn't correct.

    <br />
\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} <br />

    <br />
\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<br />
    Sorry, I was very tired when I was writing this.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 19th 2010, 05:04 PM
  2. example of irrational + irrational = rational
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 4th 2010, 03:44 AM
  3. Replies: 1
    Last Post: March 23rd 2010, 12:55 PM
  4. Replies: 2
    Last Post: January 31st 2010, 05:40 AM
  5. Replies: 7
    Last Post: January 29th 2009, 03:26 AM

Search Tags


/mathhelpforum @mathhelpforum