Results 1 to 4 of 4

Thread: Indefinite integrals

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    197

    Exclamation Indefinite integrals

    Question:
    $\displaystyle \int x^3.(2-x^2)^7 $

    $\displaystyle u=2-x^2 $
    $\displaystyle \int x^3 .u^7 $
    $\displaystyle du=-2x $
    I dont know what to do next.
    Please answer in steps ,Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by mj.alawami View Post
    Question:
    $\displaystyle \int x^3.(2-x^2)^7 $

    $\displaystyle u=2-x^2 $
    $\displaystyle \int x^3 .u^7 $
    $\displaystyle du=-2x $
    I dont know what to do next.
    Please answer in steps ,Thank you
    For this you should use integration by parts.

    Remember that $\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

    In this case, let $\displaystyle u = x^2$ so that $\displaystyle du = 2x$

    Let $\displaystyle dv = x(2 - x^2)^7$ so that $\displaystyle v = \int{x(2 - x^2)^7\,dx}$.

    $\displaystyle v = -\frac{1}{2}\int{-2x(2 - x^2)^7}$

    $\displaystyle = -\frac{1}{2}\cdot\frac{1}{8}(2 - x^2)^8$

    $\displaystyle = -\frac{1}{16}(2 - x^2)^8$.



    So $\displaystyle \int{x^3(2 - x^2)^7\,dx} = -\frac{1}{16}x^2(2 - x^2)^8 - \int{-\frac{1}{8}x(2 - x^2)^8\,dx}$

    $\displaystyle = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{16}\int{-2x(2 - x^2)^8\,dx}$

    $\displaystyle = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{16}\cdot\frac{1}{9}(2 - x^2)^9 + C$

    $\displaystyle = -\frac{1}{16}x^2(2 - x^2)^8 - \frac{1}{144}(2 - x^2)^9 + C$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    The substitution $\displaystyle u=2-x^{2}$ is itself correct, but the complete procedure is...

    $\displaystyle u=2-x^{2} \rightarrow x^{3} = (2-u)^{\frac{3}{2}}$ (1)

    $\displaystyle dx= -\frac{du}{2x}= -\frac{du}{2\cdot (2-u)^{\frac{1}{2}}}$ (2)

    ... so that the integral becomes...

    $\displaystyle - \frac{1}{2}\cdot \int u^{7} \cdot (2-u)\cdot du$ (3)

    ... that really seemes more easy than the original one...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Jan 9th 2010 at 07:13 AM. Reason: Error of sign in (3)... thanks to HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,729
    Thanks
    3010
    Quote Originally Posted by mj.alawami View Post
    Question:
    $\displaystyle \int x^3.(2-x^2)^7 $

    $\displaystyle u=2-x^2 $
    $\displaystyle \int x^3 .u^7 $
    $\displaystyle du=-2x $
    I dont know what to do next.
    Please answer in steps ,Thank you
    Again, "du= -2x" is not correct. du= -2xdx so $\displaystyle -\frac{1}{2}du= xdx$. You need to learn to write that correctly!
    Rewrite $\displaystyle \int x^2(2-x^2)^7 dx$ as $\displaystyle \int x^2(2- x^2)^7(xdx)$ Now $\displaystyle (2- x^2)^7$ becomes $\displaystyle u^7$, $\displaystyle x dx$ becomes $\displaystyle -\frac{1}{2} du$ and $\displaystyle x^2$ becomes 2- u.
    $\displaystyle \int x^3(2- x^2)^7 dx$ becomes $\displaystyle -\frac{1}{2}\int (2-u)u^7 du= -\frac{1}{2}\int (2u^7- u^8) du$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Aug 5th 2010, 11:58 AM
  2. Indefinite integrals
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Jan 9th 2010, 08:35 PM
  3. Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Apr 6th 2009, 07:05 PM
  4. Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 3rd 2008, 08:00 AM
  5. indefinite integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 30th 2006, 12:46 PM

Search Tags


/mathhelpforum @mathhelpforum