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!

Difficult Integration

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data

    ##(e^y + 1)^2 e^{-y} dx + (e^x + 1)^3 e^{-x} dy = 0##

    2. Relevant equations


    3. The attempt at a solution

    ##(e^y + 1)^2 e^{-y} dx + (e^x + 1)^3 e^{-x} dy = 0##
    ##(e^{2y} + 2 e^y + 1) e^{-y} dx + (e^{3x} + 3e^{2x} + 3e^x + 1) e^{-x} dy = 0##
    ##(e^{2y - y} + 2 e^{y - y} + e^{-y}) dx + (e^{3x - x} + 3e^{2x - x} + 3e^{x - x} + e^{-x}) dy = 0##
    ##(e^{y} + 2 e^{0} + e^{-y}) dx + (e^{2x} + 3e^{x} + 3e^{0} + e^{-x}) dy = 0##
    ##(e^{y} + 2 (1) + e^{-y}) dx + (e^{2x} + 3e^{x} + 3 (1) + e^{-x}) dy = 0##
    ##(e^{y} + 2 + e^{-y}) dx + (e^{2x} + 3e^{x} + 3 + e^{-x}) dy = 0##
    ##(e^{y} + e^{-y} + 2) dx + (e^{2x} + 3e^{x} + e^{-x} + 3) dy = 0##
    ##(e^{2x} + 3e^{x} + e^{-x} + 3) dy = - (e^{y} + e^{-y} + 2) dx##
    ## - \frac{1}{e^{y} + e^{-y} + 2} dy = \frac{1}{e^{2x} + 3e^{x} + e^{-x} + 3} dx##
    ## - \int \frac{1}{e^{y} + e^{-y} + 2} dy = \int \frac{1}{e^{2x} + 3e^{x} + e^{-x} + 3} dx##

    It is a difficult integration. What next?
     
  2. jcsd
  3. Dec 4, 2014 #2

    ShayanJ

    User Avatar
    Gold Member

    The only thing that I can see here is [itex] \frac{1}{e^y+e^{-y}+2}=\frac 1 4 \sec^2 \frac y 2[/itex].
     
  4. Dec 4, 2014 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You mean sech(), right?
     
  5. Dec 4, 2014 #4
    How do you get ##\frac{1}{e^y+e^{-y}+2}=\frac 1 4 \sec^2 \frac y 2##?

    I know that ##sech \ x = \frac{2}{e^x + e^{-x}}##.
     
    Last edited: Dec 4, 2014
  6. Dec 4, 2014 #5

    ShayanJ

    User Avatar
    Gold Member

    Sorry...Yeah, I meant the hyperbolic thing!
     
  7. Dec 4, 2014 #6
    How do you get ##\frac{1}{e^y+e^{-y}+2}=\frac 1 4 \sec^2 \frac y 2##?

    I know that ##sech \ y = \frac{2}{e^y + e^{-y}}##.
     
    Last edited: Dec 4, 2014
  8. Dec 4, 2014 #7

    ShayanJ

    User Avatar
    Gold Member

    [itex] sech x=\frac{2}{e^x+e^{-x}} -^{x=\frac{y}{2}}\rightarrow sech \frac y 2=\frac{2}{e^{\frac y 2}+e^{-\frac y 2}} \rightarrow sech^2 \frac y 2=\frac{4}{e^{y}+e^{-y}+2} [/itex]
     
    Last edited: Dec 4, 2014
  9. Dec 4, 2014 #8
    Why ##sech \ x = \frac{2}{e^x + e^{-x}} - x##?

    And why it equal to ##\frac{y}{2}##?

    The correct one is ##sech \ x = \frac{2}{e^x + e^{-x}}##, not ##\frac{2}{e^x + e^{-x}} - x##.

    What this -x variable mean?
     
  10. Dec 4, 2014 #9

    ShayanJ

    User Avatar
    Gold Member

    Oh...I meant to be a little fancy but looks like it was a bad idea. I meant take [itex]sech x=\frac{2}{e^x+e^{-x}}[/itex] and then set [itex] x=\frac y 2 [/itex].
     
  11. Dec 4, 2014 #10
    You said ##sech \ \frac{y}{2} = \frac{2}{e^{\frac{y}{2}} + e^{-\frac{y}{2}}}##
    ##= \frac{2}{e^{\frac{y}{2}} + \frac{1}{e^{\frac{y}{2}}}}##

    ##= \frac{2}{\frac{(e^{\frac{y}{2}})^2 + 1}{e^{\frac{y}{2}}}}##

    ##=\frac{2 e^{\frac{y}{2}}}{(e^{\frac{y}{2}})^2 + 1}##

    ##=\frac{2 e^{\frac{y}{2}}}{(e^{\frac{y}{2}}. \ e^{\frac{y}{2}}) + 1}##

    ##=\frac{2 e^{\frac{y}{2}}}{e^{\frac{y}{2} + \frac{y}{2}} + 1}##

    ##=\frac{2 e^{\frac{y}{2}}}{e^{\frac{y + y}{2}} + 1}##

    ##=\frac{2 e^{\frac{y}{2}}}{e^{\frac{2y}{2}} + 1}##

    ##=\frac{2 e^{\frac{y}{2}}}{e^{y} + 1}##

    Please explain. I am really confuse.
     
  12. Dec 4, 2014 #11

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    When you write this

    exdx/(ex + 1)3 = - eydy/(ey + 1)2

    isn't each side easily recognised as derivative of something?
     
  13. Dec 4, 2014 #12

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Or if you don't recognize the derivatives, try the obvious substitutions.
     
  14. Dec 5, 2014 #13
    I still don't get it. Give me more clue please.
     
  15. Dec 5, 2014 #14

    Mark44

    Staff: Mentor

    Vela gave you a clue in post #12. Try it.
     
  16. Dec 5, 2014 #15
    ##\frac{e^x dx}{(e^x + 1)^3} = \frac{-e^y dy}{(e^y + 1)^2}##

    I shall integrate the left side (the x variable) first. If using the substitution method, which one should be substituted?
    Let ##u = e^x## or ##u = e^x + 1##?
     
  17. Dec 5, 2014 #16

    Mark44

    Staff: Mentor

    Try them both. You should see that one is slightly better than the other.
     
  18. Dec 5, 2014 #17
    ##\frac{e^x dx}{(e^x + 1)^3} = \frac{-e^y dy}{(e^y + 1)^2}##

    (1)
    Let ##u = e^x + 1## then ##\frac{du}{dx} = e^x## or ##du = e^x dx##

    ##\int \frac{e^x dx}{(e^x + 1)^3}##
    ##= \int \frac{1}{u^3}du##
    ##= \ln |u^3| + c##
    ##= \ln (e^x + 1)^3 + c##

    (2)
    Let ##u = e^y + 1## then ##\frac{du}{dy} = e^y## or ##du = e^y dy##

    ##\int \frac{-e^y dy}{(e^y + 1)^2}##
    ##= - \int \frac{e^y dy}{(e^y + 1)^2}##
    ##= - \int \frac{1}{u^2}du##
    ##= - \ln |u^2| + c##
    ##= - \ln (e^y + 1)^2 + c##

    From (1) and (2), I get:

    ##\int \frac{e^x dx}{(e^x + 1)^3} = \int \frac{-e^y dy}{(e^y + 1)^2}##
    ##\ln (e^x + 1)^3 + c = - \ln (e^y + 1)^2 + c##

    Is it correct?
     
  19. Dec 5, 2014 #18

    Zondrina

    User Avatar
    Homework Helper

    That is certainly not correct. What is:

    $$\int u^{-3} \space du$$
     
  20. Dec 5, 2014 #19

    Mark44

    Staff: Mentor

    Good catch, Zondrina. I totally missed that he was using this integration "rule" (in quotes because it's not an integration rule):
    ##\int \frac{dx}{f(x)} = ln|f(x)| + C##
     
  21. Dec 5, 2014 #20
    Thank you for the correction.

    Let I correct the mistake:

    ##\frac{e^x dx}{(e^x + 1)^3} = \frac{-e^y dy}{(e^y + 1)^2}##

    (1)
    Let ##u = e^x + 1## then ##\frac{du}{dx} = e^x## or ##du = e^x dx##

    ##\int \frac{e^x dx}{(e^x + 1)^3}##
    ##= \int \frac{1}{u^3}du##
    ##= \int u^{-3} du##
    ##= \frac{1}{-3+1}u^{-3+1} + c##
    ##= \frac{1}{-2}u^{-2} + c##
    ##= - \frac{1}{2}(e^x + 1)^{-2} + c##

    (2)
    Let ##u = e^y + 1## then ##\frac{du}{dy} = e^y## or ##du = e^y dy##

    ##\int \frac{-e^y dy}{(e^y + 1)^2}##
    ##= - \int \frac{e^y dy}{(e^y + 1)^2}##
    ##= - \int \frac{1}{u^2}du##
    ##= - \int u^{-2} du##
    ##= - \frac{1}{-2+1} u^{-2+1} + c##
    ##= - \frac{1}{-1} u^{-1} + c##
    ##= u^{-1} + c##
    ##= (e^y + 1)^{-1} + c##

    From (1) and (2), I get:

    ##\int \frac{e^x dx}{(e^x + 1)^3} = \int \frac{-e^y dy}{(e^y + 1)^2}##
    ##- \frac{1}{2}(e^x + 1)^{-2} + c = (e^y + 1)^{-1} + c##

    Is it correct?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Difficult Integration
  1. Difficult integral (Replies: 4)

  2. Difficult Integral (Replies: 10)

  3. Difficult Integral (Replies: 31)

  4. Difficult integral (Replies: 7)

  5. Difficult integral. (Replies: 5)

Loading...