Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Plato

Thread: integration by parts

  1. #1
    Newbie
    Joined
    Mar 2013
    From
    USA
    Posts
    22

    integration by parts

    I am working on a integration by parts problem.

    $\displaystyle \int tan^{-1}(4t)dt$

    Setting $\displaystyle u=tan^{-1}(4t)$, then $\displaystyle dv=dt$, $\displaystyle dv=\frac{1}{1+(4t)^2}$ and $\displaystyle v=t$

    Plugging into the equation, we get:

    $\displaystyle tan^{-1}(4t)(t)-\int \frac{4t}{1+(4t)^2}dt$

    I am good with all of that, except the 4 in the numerator of the integrand. I don't see where that comes from, since v=t. With the 4, it does lead to the correct answer.

    Can someone explain?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1

    Re: integration by parts

    Quote Originally Posted by baldysm View Post
    I am working on a integration by parts problem.

    $\displaystyle \int tan^{-1}(4t)dt$
    Look at this webpage.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,779
    Thanks
    3028

    Re: integration by parts

    Quote Originally Posted by baldysm View Post
    I am working on a integration by parts problem.

    $\displaystyle \int tan^{-1}(4t)dt$

    Setting $\displaystyle u=tan^{-1}(4t)$, then $\displaystyle dv=dt$, $\displaystyle dv=\frac{1}{1+(4t)^2}$ and $\displaystyle v=t$
    You have a lot of typos that make this very difficult to read. You mean
    $\displaystyle u= tan^{-1}(4t)$ and $\displaystyle dv= dt$. Then $\displaystyle du= \frac{4}{1+ (4t)^2} dt$ and $\displaystyle v= t$. An error that I think is NOT a typo is that you have $\displaystyle du= \frac{1}{1+ (4t)^2}$ without the "4" in the numerator. You need that because of the "chain rule". To differentiate $\displaystyle u= tan^{-1}(4t)$, let y= 4t so that $\displaystyle u= tan^{-1}(y)$. Then $\displaystyle \frac{du}{dy}= \frac{1}{1+ y^2}$ and $\displaystyle \frac{dy}{dx}= 4$. By the chain rule, $\displaystyle \frac{du}{dx}= \frac{du}{dy}\frac{dy}{dx}= \frac{1}{1+ u^2}(4)= \frac{4}{1+ 16t^2}$.

    Plugging into the equation, we get:

    $\displaystyle tan^{-1}(4t)(t)-\int \frac{4t}{1+(4t)^2}dt$

    I am good with all of that, except the 4 in the numerator of the integrand. I don't see where that comes from, since v=t. With the 4, it does lead to the correct answer.

    Can someone explain?
    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. Integration by parts...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 8th 2011, 12:01 PM
  3. Replies: 8
    Last Post: Sep 2nd 2010, 12:27 PM
  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