Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By skeeter

Thread: Integration by parts

  1. #1
    Junior Member
    Joined
    Nov 2011
    Posts
    33
    Thanks
    1

    Integration by parts

    Hello!

    I am trying to solve the following indefinite integral by using integration by parts (as is recommended by the textbook): $\displaystyle \int{(e^x\sin{x})dx}$.

    However, I am stuck in the same circle over and over again.

    $\displaystyle U = e^x$ and $\displaystyle dU = (e^x)dx$.
    $\displaystyle dV = (\sin{x})dx$ and $\displaystyle V = -\cos{x}$.

    So: $\displaystyle \int{(U)dV} = UV - \int{(V)dU}$ gives $\displaystyle \int{(e^x\sin{x})dx} = -e^x\cos{x} - \int{(-e^x\cos{x})dx}$.

    Resulting in the same problem when trying to find $\displaystyle \int{(-e^x\cos{x})dx}$.

    Even when I define $\displaystyle U$ and $\displaystyle dV$ the other way around, I am still stuck with $\displaystyle e^x$ and either $\displaystyle \sin{x}$ or $\displaystyle \cos{x}$.

    How do I do this? Please help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702

    Re: Integration by parts

    $\displaystyle \int e^x \sin{x} \, dx$

    $\displaystyle u = \sin{x} \, , \, dv = e^x \, dx$

    $\displaystyle du = \cos{x} \, dx \, , \, v = e^x$

    $\displaystyle \int e^x \sin{x} \, dx = e^x \sin{x} - \int e^x \cos{x} \, dx$

    parts again with the last integral expression ...

    $\displaystyle u = \cos{x} \, , \, dv = e^x \, dx$

    $\displaystyle du = -\sin{x} \, dx \, , \, v = e^x$

    $\displaystyle \int e^x \sin{x} \, dx = e^x \sin{x} - \left[e^x \cos{x} - \int e^x (-\sin{x}) \. dx \right]$

    $\displaystyle \int e^x \sin{x} \, dx = e^x \sin{x} - e^x \cos{x} - \int e^x \sin{x} \, dx$

    $\displaystyle 2\int e^x \sin{x} \, dx = e^x \sin{x} - e^x \cos{x}$

    $\displaystyle \int e^x \sin{x} \, dx = \frac{e^x(\sin{x}-\cos{x})}{2} + C$
    Thanks from Lotte1990
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Jan 11th 2012, 02:30 PM
  2. Replies: 8
    Last Post: Sep 2nd 2010, 12:27 PM
  3. U-sub or integration by parts?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 25th 2010, 03:34 AM
  4. Replies: 0
    Last Post: Apr 23rd 2010, 03:01 PM
  5. Replies: 1
    Last Post: Feb 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum