1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rewriting integrals

  1. Jan 28, 2008 #1
    How does one evaluate [tex]\int (arctan(pi*x) - arctan(x))dx[/tex] from 0 to 2 by rewriting the integrand as an integral?
    Last edited: Jan 28, 2008
  2. jcsd
  3. Jan 28, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    The point is that you have F(b)- F(a) which can be written as
    [tex]\int_a^b f(x) dx[/itex] where f is the derivative of F. The derivative of arctan(x) is
    [tex]\frac{x^2+ 1}[/itex] so this could be
    [tex]\int_0^2 \int_x^{\pi x} \frac{1}{t^2+ 1} dt dx[/itex]

    Now see what happens if you reverse the order of integration:
    For every x, t goes from x to [itex]\pi x[/itex] so boundaries on the region on which you are integrating are t= x, t= [itex]\pi x[/itex], and x= 2 (x= 0 just gives the intersection of the two straight lines).

    If you reverse the order of integration, you will need to take t going from 0 to [itex]2\pi[/itex]. The limits of integration for x are a bit more complicated. The left end will be at [itex]t= \pi x[/itex] or [itex]x= t/\pi[/itex]. For t between 0 and 2, the right end will be t= x or x= t, but for t between 2 and [itex]2\pi[/itex] the right end is at x= 2. The integral is
    [tex]\int_{t=0}^2 \int_{x= t/\pi}^t \frac{1}{1+ t^2} dx dt+ \int_{t= 2}^{2\pi} \int_{x= t/\pi}^2 \frac{1}{1+t^2} dx dt[/itex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Rewriting integrals
  1. On Integration (Replies: 4)

  2. An integral (Replies: 2)