Results 1 to 3 of 3

Math Help - Integration by Substitution

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    107

    Integration by Substitution

    Hey everyone. I am having trouble with this problem:

    \int \frac{x^2}{\sqrt{1-x}}dx

    I can only use substitution to do this. I have no idea what to do. I tried rewriting it in to something that looks integrable using substitution but everything has failed. Here are some of my attempts at rewriting it:

    \sqrt{\frac{x^4}{1-x}}

    \frac{x^2\sqrt{1+x}}{\sqrt{1-x^2}}

    \frac{x^2\sqrt{1-x}}{1-x}

    \frac{\sqrt{x^4-x^5}}{1-x}

    And non of them are integrable by substitution in my eyes. Can someone help me out?

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Let y = 1-x, then you have \int\frac{x^2}{\sqrt{1-x}}\;{dx} = -\int\frac{(1-y)^2}{\sqrt{y}}\;{dy}. From there it's
    just routine: expand the numerator, divide each term by \sqrt{y} then integrate.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    773
    Hello, lilaziz1!

    \displaystyle \int \frac{x^2}{\sqrt{1-x}}\,dx

    "Substitution" usually means to let \,u equal something already in the problem.


    Let u\,=\,\sqrt{1-x} \quad\Rightarrow\quad u^2 \,=\,1-x

    Then: . x \,=\,1-u^2 \quad\Rightarrow\quad dx \,=\,-2u\,du


    Substitute: . \displaystyle \int\frac{(1-u^2)^2}{u}\,(-2u\,du) \;=\;-2\int(1-u^2)^2\,du

    . . \displaystyle =\;-2\int \left(1-2u^2+u^4\right)\,du \;=\;-2\left(u - \tfrac{2}{3}u^3 +\tfrac{1}{5}u^5\right) + C

    . . =\;-\frac{2}{15}u(15 - 10u^2 + 3u^4) + C


    Back-substitute: . -\frac{2}{15}\sqrt{1-x}\bigg[15 - 10(1-x) + 3(1-x)^2\bigg] + C

    . . =\;-\frac{2}{15}\sqrt{1-x}\bigg[15 - 10 + 10x + 3 - 6x + 3x^2\bigg] + C

    . . =\;-\frac{2}{15}\sqrt{1-x}\,(3x^2 + 4x + 8) + C

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration by substitution
    Posted in the Calculus Forum
    Replies: 13
    Last Post: December 26th 2010, 11:24 PM
  2. Integration by substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 9th 2009, 06:08 PM
  3. integration by substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 5th 2008, 01:38 PM
  4. Integration by Substitution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 27th 2008, 06:14 PM
  5. substitution integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 6th 2007, 08:51 PM

Search Tags


/mathhelpforum @mathhelpforum