Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook