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!

Homework Help: Please help:Calc II-Integration

  1. Sep 18, 2009 #1
    This is my first posting on the site. I read the directions, I hope I follow all applicable rules. In advance, thank you for your help.

    1. The problem statement, all variables and given/known data
    Find indefinite integral:
    Variables (X)

    2. Relevant equations

    3. The attempt at a solution
    DU=eCot(X) * -Csc2(X)
    Pyth Iden Csc2(X)=1 + Tan2(X)
    I keep going back to:
    [tex]\int[/tex][Csc2(Something)]=-Cot(Something) I just can't figure out the relation.

    1. The problem statement, all variables and given/known data
    Find indef. Int.
    Variables: X
    [tex]\int[/tex] [tex]\frac{5}{3e^x -2}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    5[tex]\int[/tex][tex]\frac{1+3e^x -3e^x}{3e^x -2}[/tex]
    5[tex]\int[/tex][tex]\frac{1+3e^x}{3e^x -2}[/tex] - 5[tex]\int[/tex][tex]\frac{DU}{U}[/tex]

    Then I can't Integrate the first fraction.

    Then tried conjugate:
    5[tex]\int[/tex][tex]\frac{9e^{2x} -3e^x -2}{9e^{2x} -4}[/tex]

    Here the den. is similar to the ArcTrig integrals, however the signs are wrong and there is no sq.rt.
    Last edited: Sep 18, 2009
  2. jcsd
  3. Sep 18, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    Hi The1BigJones, welcome to PF!

    There's no need to use any trig identities here. If [itex]u=e^{\cot x}[/itex], then [itex]du=-\csc^2(x) e^{\cot x} dx[/itex] (You left out the [itex]dx[/itex] from your expression)....[itex]du[/itex] looks an awful lot like your integrand to me!:wink:

    Okay, let's call this integral [itex]I[/itex]

    [tex]I\equiv\int\frac{5}{3e^x -2}dx[/tex]



    So now you have,

    [tex]\int\frac{5}{3e^x -2}dx=5\int\left(1+\frac{3}{3e^x-2}\right)dx -5\ln|3e^x-2|[/tex]

    Break the integral into two pieces, and substitute [itex]I[/itex] in for [itex]\int\frac{5}{3e^x -2}dx[/itex] on both sides of the equation, and then solve for [itex]I[/itex] algebraically.
  4. Sep 19, 2009 #3


    User Avatar

    If you substitute [tex]x=\ln{t}, dx=\frac{dt}{t}[/tex], then you'll end up with [tex]\int \frac{5}{t(3t-2)}\,dt[/tex]. From there you can apply partial fraction decomposition.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook