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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)
    Problem:
    [tex]\int[/tex]Csc2(eCot(X))DX

    2. Relevant equations
    None:


    3. The attempt at a solution
    U=eCot(X)
    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
    None

    3. The attempt at a solution
    U=3ex-2
    DU=3ex

    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

    gabbagabbahey

    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]

    Hint:

    [tex]\frac{1+3e^x}{3e^x-2}=\frac{3e^x-2+3}{3e^x-2}=1+\frac{3}{3e^x-2}[/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

    zcd

    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.
     
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