Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - Improper Integrals!!!!!!

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    87

    Improper Integrals!!!!!!

    I need help integrating from 0 to 1, the integral of (41) all over (square root of 1-x^2). I think that this integral converges, but I'm not sure if I'm evaluating the integral right, I started by putting the 41 in front of the integral and maybe using trig substitution?! Ah please help me, I am out of it, with math!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    \int{\frac{dx}{\sqrt{1-x^{2}}}}=\arcsin x+k, which is a standard integral.

    Now if you wanna prove such result, make the substitution x=\sin\varphi,\,-\frac\pi2\le\varphi\le\frac\pi2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    87
    Okay! I used x=sintheta and so I got (1) over the square root of (1-(sintheta squared) squared. That looks kind of weird, but then I got rid of the square root and got (1) over (1-sin(theta)^2). That is the same as (1) over cos(theta)^2, which is the same thing as secant squared, and then integrating that gets you tangent I think. So, I'm that far. I think I have to use the triangle, to change theta back to x. I think that is opposite over adjacent, which is (x) over (the square root of 1-x^2). I'm not sure where to go from there.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by elocin View Post
    Okay! I used x=sintheta and so I got (1) over the square root of (1-(sintheta squared) squared.
    There's your mistake. What you have written is
    \int \frac{1}{ \left ( \sqrt{1 - sin^2(\theta)} \right ) ^2}~dx

    You have an extra "squared" in there. It should be
    \int \frac{1}{\sqrt{1 - sin^2(\theta)}}~dx

    Also,
    x = sin(\theta) \implies dx = cos(\theta)~d \theta
    You need to replace the differential element as well.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2008
    Posts
    87
    Hmm, okay So I have the integral of (1) over the square root of 1-sintheta squared and cos(theta) dtheta. I simplified that to be (1) over the square root cosine squared, cosine theta d theta. So I simplified that to be (1) over cosine theta, cosine theta d theta. I canceled out a cosine theta, and therefore got 1 dtheta. Can that be right? Integrating that, I would get just theta?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by elocin View Post
    Hmm, okay So I have the integral of (1) over the square root of 1-sintheta squared and cos(theta) dtheta. I simplified that to be (1) over the square root cosine squared, cosine theta d theta. So I simplified that to be (1) over cosine theta, cosine theta d theta. I canceled out a cosine theta, and therefore got 1 dtheta. Can that be right? Integrating that, I would get just theta?
    Yes. You should get \theta+K. But what is theta? Recall that you made the substitution x=\sin(\theta)...

    --Chris
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jun 2008
    Posts
    87
    I'm not sure, I know I would use the sine triangle if I had tangent or cosine or something, but all I have is theta!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by elocin View Post
    I'm not sure, I know I would use the sine triangle if I had tangent or cosine or something, but all I have is theta!
    Recall that the integration variable is a "dummy" variable.
    \int d \theta = \theta + C

    is the same process as
    \int dx = x + C

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by elocin View Post
    I'm not sure, I know I would use the sine triangle if I had tangent or cosine or something, but all I have is theta!
    Well, we made the substitution x=\sin(\theta). Solving for theta, we get \sin^{-1}(x)=\sin^{-1}(\sin(\theta)) \implies \color{red}\boxed{\theta=\sin^{-1}(x)}.

    So instead of \theta+k as our answer, it would be \sin^{-1}(x)+k.

    Hope this makes sense!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jun 2008
    Posts
    87
    So integration just gives you x? multiplied by 41? I'm confused!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Jun 2008
    Posts
    87
    Oh okay!! Thank you!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jun 2008
    Posts
    87
    So I got 41arcsin(x), so would I just plug in 0 and 1? That gives me 41arcsin(1), since arcsin(0) =0.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by elocin View Post
    So I got 41arcsin(x), so would I just plug in 0 and 1? That gives me 41arcsin(1), since arcsin(0) =0.
    Yes. Note that \sin^{-1}(1)=\frac{\pi}{2}. So your answer is 42\bigg[\frac{\pi}{2}\bigg]=\color{red}\boxed{21\pi}

    Hope this makes sense!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Jun 2008
    Posts
    87
    Well the original integral was integrate from 0 to 1, the integral of (41) over (square root of 1-x^2), so wouldn't it be 41pi/2? I'm sorry, I just don't know where the 42 came from!
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by elocin View Post
    Well the original integral was integrate from 0 to 1, the integral of (41) over (square root of 1-x^2), so wouldn't it be 41pi/2? I'm sorry, I just don't know where the 42 came from!
    It was a typo. It should be \frac{41\pi}{2}

    Sorry about that!
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Improper Integrals
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: February 15th 2011, 12:09 AM
  2. improper integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 7th 2009, 05:06 AM
  3. Improper Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 17th 2009, 08:29 PM
  4. Improper Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 5th 2008, 11:57 AM
  5. Improper Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 24th 2007, 05:05 PM

Search Tags


/mathhelpforum @mathhelpforum