Results 1 to 5 of 5

Math Help - Integration question

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    6

    Integration question

    Prove that \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)

    Hints would be appreciated. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    185
    Thanks
    13
    Quote Originally Posted by anthill View Post
    Prove that \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)

    Hints would be appreciated. Thanks.
    substitute 1-x^2=t^2 and express x^2 in denominator using that substitution. You should get a rational function in variable t.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by anthill View Post
    Prove that \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)

    Hints would be appreciated. Thanks.
    \displaystyle \int \frac{\sqrt{1-x^2} \cdot \sqrt{1-x^2}}{(1+x^2)\cdot \sqrt{1-x^2}} dx

    \displaystyle \int \frac{-1 -x^2 }{(1+x^2) \cdot \sqrt{1-x^2}}  + \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx

    \displaystyle \int \frac{-1 -x^2 }{(1+x^2) \cdot \sqrt{1-x^2}} dx +\int   \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx

    \displaystyle \int \frac{-1 }{\sqrt{1-x^2}} dx +\int   \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx

    first integration sub
    \sin u  = x
    in the second sub
    \displaystyle \tan u = \frac{\sqrt{2} x }{\sqrt{1-x^2}}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Let  x = \sin(y)

    dx = \cos(y)~dy


    The integral

     = \int_0^{\frac{\pi}{2}} \frac{\cos^2(y)}{ 1+ \sin^2(y) }~dy

     =  \int_0^{\frac{\pi}{2}} \frac{\cos^2(y)}{ \cos^2(y) + 2\sin^2(y) }~dy

     =  \int_0^{\frac{\pi}{2}} \frac{dy}{ 1+ 2\tan^2(y) }

    Consider  1 = \sec^2(y) - \tan^2(y) - 1/2 + 1/2 thus we have

     \frac{1}{2}  = \sec^2(y) - \frac{1}{2} ( 1+ 2\tan^2(y) )


    The integral

     = 2 \int_0^{\frac{\pi}{2}} \frac{\sec^2(y) }{ 1+ 2\tan^2(y) }~dy - \int_0^{\frac{\pi}{2}} ~dy


     = \left[  \sqrt{2} \tan^{-1}[\sqrt{2}\tan(y)] - y \right]_0^{\frac{\pi}{2}}


     = \frac{\pi}{2}( \sqrt{2} - 1 )
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: July 21st 2010, 06:20 PM
  2. Integration question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 5th 2010, 11:19 AM
  3. i need help in integration question?
    Posted in the Calculus Forum
    Replies: 11
    Last Post: January 24th 2010, 07:03 PM
  4. Integration question
    Posted in the Calculus Forum
    Replies: 12
    Last Post: December 7th 2009, 03:49 PM
  5. integration question help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 17th 2008, 06:48 PM

Search Tags


/mathhelpforum @mathhelpforum