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!

Solving integral with substitution

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data
    find the integral from 0 to ln2 of (e^x-1)/(e^x+1)


    2. Relevant equations



    3. The attempt at a solution
    I really didn't know what to do, but the book had a hint suggesting i multiply the numerator and denominator by e^-x and then use an appropriate substitution..
    [((e^x)-1)*(e^-x)]/[((e^x)+1)*(e^-x)]
    = (1-(e^-x))/(1+(e^-x))
    = (1-(e^x))/(1+(e^x))
    but now i can't think of an appropriate substitution. help?
     
  2. jcsd
  3. Aug 10, 2011 #2
    How did you get from:
    [itex]\frac{1-e^{-x}}{1+e^{-x}}[/itex]
    To:
    [itex]\frac{1-e^{x}}{1+e^{x}}[/itex]
     
  4. Aug 10, 2011 #3
    apparently i just made up my own rules of calculus.. you're right, i can't do that.
    so i'm just at
    [1-e^(-x)]/[1+e^(-x)]
     
  5. Aug 10, 2011 #4
    Hahah I don't know how your book wants you to do it but try splitting it into two integrals.
     
  6. Aug 10, 2011 #5
    how can i split that?
     
  7. Aug 10, 2011 #6
    Into two fractions - you have addition on top - and then attack them both separately.
     
  8. Aug 10, 2011 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The first thing I would do is divide:
    [tex]\frac{e^x- 1}{e^x+ 1}= 1- \frac{2}{e^x+ 1}[/tex]

    Then, for the second integral, let [itex]u= e^x+ 1[/itex]
     
  9. Aug 10, 2011 #8
    lol... what did you divide by to get that?
    and also, if i use u=e^x, then i get du=e^xdx, and i don't have anything to cancel the extra e^x with..?
     
  10. Aug 10, 2011 #9
    I'm not sure what multiplying top & bottom by e^-x does for you, but if you multiply top & bottom by e^(-x/2), you'll probably recognize it as the hyperbolic tangent of x/2.
     
  11. Aug 10, 2011 #10
    You can use long division to divide them, similar to normal division of numbers. Or you can write it as [tex]\frac{e^x - 1}{e^x + 1} = \frac{e^x + 1 - 2}{e^x + 1} = \frac{e^x + 1}{e^x + 1} - \frac{2}{e^x + 1} = 1 - \frac{2}{e^x + 1}[/tex]
    If you want to follow Hall's suggestion, you should multiply the fraction by e-x/e-x and then use a different substitution.

    I would follow c.dube's suggestion by splitting the fraction in two [tex]\frac{e^x - 1}{e^x + 1} = \frac{e^x}{e^x + 1} - \frac{1}{e^x + 1}[/tex]The first one has an easy substitution, but you'll want to multiply the second by e-x/e-x before using a substitution.
     
  12. Aug 10, 2011 #11
    ookay so i ended up with
    ln(e^(x)+1)+ln(e^(-x)+1)... is that right?
     
  13. Aug 10, 2011 #12
    ookay so i ended up with
    ln(e^(x)+1)+ln(e^(-x)+1) between 0 and ln2
    = ln((e^x+1)/(e^-x)+1)) between 0 and ln2
    = ln ((e^ln2+1)/(e^-ln2+1))-ln((e^0+1)/e^0+1))
    = ln(3/-1)-ln(1)
    = ln(-3)

    .. which is very wrong, obviously.
    where did i go wrong?
     
  14. Aug 10, 2011 #13
    The two logs are being added, so you need to multiply their arguments and not divide them.
     
  15. Aug 10, 2011 #14
    ahhh! of course!
    so now i get

    ln((e^x+1)*(e^-x+1))
    = [ln((e^ln2+1)*(e^-ln2+1))]-[ln(e^0+1)*(e^0+1)]
    = ln(-3)-ln(4)
    = ln(-3/4)

    closer? but still not right...?
     
  16. Aug 10, 2011 #15
    e^-ln2 ≠ -2
    Make sure you use the rules of logs and exponents correctly when you simplify.
     
  17. Aug 10, 2011 #16
    Look at the part [itex]e^{-\ln2}[/itex] and how you could re-express that. Then run with it.
    EDIT: Haha as Bohrok said.
     
  18. Aug 10, 2011 #17
    e^-ln2= (1/2)??
     
  19. Aug 10, 2011 #18
    Yes! :smile:
     
  20. Aug 10, 2011 #19
    got it! :):)
     
    Last edited: Aug 10, 2011
  21. Aug 12, 2011 #20
    You guys are making this much too difficult! Note that the integrand is tanh(x/2) which is in turn sinh(x/2)/cosh(x/2). Substitute u = sinh(x/2) and recall that the derivative of sinh is cosh. The integrand then becomes 1/u, which integrates to ln(u). Resolve the substitution and voila.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook