Results 1 to 3 of 3

Math Help - Integrate by completing the square.

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    4

    Integrate by completing the square.

    Integrate by completing the square.-factoring.jpg
    Last edited by mr fantastic; August 16th 2010 at 10:06 PM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    768
    Hello, roza!

    By completing the square, show that:

    . . \displaystyle \int^{\frac{1}{2}}_0 \frac{dx}{x^2-x+1} \;=\;\frac{\pi}{3\sqrt{3}}

    Complete the square: . (x^2-x+\frac{1}{4}) + 1 - \frac{1}{4}

    . . . . . . =\;(x-\frac{1}{2})^2 + \frac{3}{4} \;=\; (x - \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2


    We have: . \displaystyle \int^{\frac{1}{2}}_0 \frac{dx}{\left(x-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}


    Let: . x - \frac{1}{2} \:=\:\frac{\sqrt{3}}{2}\tan\theta \quad\Rightarrow\quad dx \:=\:\frac{\sqrt{3}}{2}\sec^2\!\theta\,d\theta

    . . and: . \left(x-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \;=\; \frac{3}{4}\sec^2\!\theta


    Substitute: . \displaystyle \int\dfrac{\frac{\sqrt{3}}{2}\sec^2\!\theta\,d\the  ta}{\frac{3}{4}\sec^2\!\theta} \;=\; \frac{2}{\sqrt{3}}\!\int d\theta \;=\;\frac{2}{\sqrt{3}}\,\theta + C


    Back-substitute: . \tan\theta \:=\:\dfrac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \;=\;\dfrac{2x-1}{\sqrt{3}}


    We have: . \dfrac{2}{\sqrt{3}}\,\arctan\left(\dfrac{2x-1}{\sqrt{3}}\right)\,\bigg]^{\frac{1}{2}}_0

    . . . . =\; \dfrac{2}{\sqrt{3}}\,\bigg[\arctan 0 - \arctan\left(\text{-}\dfrac{1}{\sqrt{3}}\right)\bigg]

    . . . . =\;\dfrac{2}{\sqrt{3}}\,\bigg[0 - \left(\text{-}\dfrac{\pi}{6}\right)\bigg]

    . . . . =\;\dfrac{\pi}{3\sqrt{3}}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    \int{0}{1\2}\frac{dx}{(x^2 - x + 1)}

    Write x^2 - x + 1 as x^2 - x + 1/4 + 3/4 = (x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2

    This reduces the given integration to a standard form. Now solve it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Completing the Square
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 17th 2010, 12:33 AM
  2. Completing the Square
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 6th 2009, 05:23 PM
  3. Completing the square
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 25th 2009, 09:20 AM
  4. Help me with Completing the square
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 20th 2008, 03:04 PM
  5. completing the square
    Posted in the Algebra Forum
    Replies: 4
    Last Post: July 11th 2008, 08:57 PM

Search Tags


/mathhelpforum @mathhelpforum