1. Limited time only! Sign up for a free 30min personal 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!

Nasty integral - which piece of technology to trust?

  1. Sep 3, 2010 #1
    Nasty integral -- which piece of technology to trust?

    1. The problem statement, all variables and given/known data

    indefinite integral: e^x / (e^2x + 9) dx
    Technology is allowed to solve it.

    2. Relevant equations




    3. The attempt at a solution

    My trusty TI-89 says (pi/540)arctan((e^3x)/3)

    while two different online integral calculators came up with (1/3log(e))*arctan((e^3x)/3)

    What do you think? Which is right/should I trust? Should I just do it by hand? If so, what method do you recommend?

    Thanks
     
  2. jcsd
  3. Sep 3, 2010 #2
    Re: Nasty integral -- which piece of technology to trust?

    u=e^x
     
  4. Sep 3, 2010 #3
    Re: Nasty integral -- which piece of technology to trust?

    Hmm, I don't know if it's just by foggy memory of Calc II but that substitution doesn't seem to help it at all.
     
  5. Sep 3, 2010 #4
    Re: Nasty integral -- which piece of technology to trust?

    Make the subst [itex]e^{x} = 3 u[/tex] and your integral will be converted to a table one. none of the answers is correct.
     
  6. Sep 3, 2010 #5

    statdad

    User Avatar
    Homework Helper

    Re: Nasty integral -- which piece of technology to trust?

    What relationship is there between [tex] e^{2x} [/tex] and [tex] e^x [/tex]? That should help with the integral and the substitution.
     
  7. Sep 3, 2010 #6
    Re: Nasty integral -- which piece of technology to trust?

    Thanks you two, I am starting to get it, but I am running into trouble.

    So far, 3u=e^x, du=?dx -- should I take du/dx of e^x/3 and get du = e^x dx? But if I do that then I get an e^x in my integral below.

    integral: 3u / (9u^2 + 9)
     
  8. Sep 3, 2010 #7
    Re: Nasty integral -- which piece of technology to trust?

    what is du?
     
  9. Sep 3, 2010 #8
    Re: Nasty integral -- which piece of technology to trust?

    The derivative of u. I thought I had to take the derivative of my subsitution and solve for dx to sub that back into my integral.
     
  10. Sep 3, 2010 #9
    Re: Nasty integral -- which piece of technology to trust?

    Write what you get for du.
     
  11. Sep 3, 2010 #10

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Nasty integral -- which piece of technology to trust?

    I think your trusty TI-89 calculated exactly what you told it to calculate, rather than what you meant to calculate. Is it in degree mode, perchance, instead of radian mode?
     
  12. Sep 3, 2010 #11
    Re: Nasty integral -- which piece of technology to trust?

    yes :)
     
  13. Sep 3, 2010 #12
    Re: Nasty integral -- which piece of technology to trust?

    so I got du = (e^x)/3 dx. What do I do with this expression? I thought I was supposed to solve for dx, and plug that into my integral with my u's. Is that not right?

    integral: 3u / (9u^2 + 9) * 3 / (e^x) du
     
  14. Sep 3, 2010 #13
    Re: Nasty integral -- which piece of technology to trust?

    Hurkyl, it was in radian mode.
     
  15. Sep 3, 2010 #14
    Re: Nasty integral -- which piece of technology to trust?

    Oh, so then the default output of the arctangent function for TI-89 is in degrees.
     
  16. Sep 3, 2010 #15
    Re: Nasty integral -- which piece of technology to trust?

    Oh wow, I just did it in degree mode and it produced the answer consistent with the back of the book.
     
  17. Sep 3, 2010 #16
    Re: Nasty integral -- which piece of technology to trust?

    Thanks Statdad, Dickfore and Hurkyl, so very much!
     
  18. Sep 3, 2010 #17
    Re: Nasty integral -- which piece of technology to trust?

    Oh and I finally figured out the substitution thing. I realized I didn't need to sub 3u into the numerator
     
  19. Sep 4, 2010 #18
    Re: Nasty integral -- which piece of technology to trust?

    u=e^x
    du=e^xdx
    Your integral becomes
    1/(u^2 +9) which is the antiderievative of (1/3)arctan(u/3).....
     
    Last edited: Sep 4, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Nasty integral - which piece of technology to trust?
  1. Nasty Integrals (Replies: 3)

  2. Solving nasty integral (Replies: 16)

Loading...