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Integration by parts HELP

  1. Feb 27, 2008 #1
    Integration by parts HELP!!

    Hey, im working on some calculus and im having some trouble with the last few integration by parts problems. I got the first couple, and i grasp the concept of integration by parts but for some reason I just cant figure these 3. Any help would be greatly appreciated.

    1. The integral of (e^2x)sin3xdx

    2. The integral of cos(3x+1)cos(5x+6)dx

    3. The integral of x²ln²xdx

    P.S> Feel free to ask if you cant read the questions, I tried to write them out the best I could.

    Thanks in advance,

    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2


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    Homework Helper

    1) u=e^2x ; dv=sin3x dx

    du=2e^2x dx ; v=[itex]\frac{-1}{3}cos3x[/itex]

    [tex]\int e^{2x}sin3x dx= \frac{-e^{2x}}{3}cos3x-\int \frac{-2e^{2x}}{3}cos3x dx[/tex]

    repeat in the same fashion
  4. Feb 27, 2008 #3
    3) [tex]\int x^{2}ln^{2}xdx[/tex] now let
    [tex]u=ln^{2}x \ so \ du=\frac{2lnx}{x} \ , \ v=\int x^{2}dx=\frac{x^{3}}{3}[/tex] now:
    [tex]\frac{x^{3}}{3}ln^{2}x-\int \frac{2lnx \ x^{3}}{3x}dx=\frac{x^{3}}{3}ln^{2}x-\frac{2}{3}\int x^{2}lnxdx[/tex] now repeat integraion by part also for

    [tex]\int x^{2}lnx dx[/tex], and i think you will get the answer!
  5. Feb 27, 2008 #4
    i am going to give u a hint on the 2) also
    [tex]I=\int cos(3x+1)cos(5x+6)dx[/tex] now let[tex]u=cos(3x+1) \ => \ du=-3sin(3x+1) \ and \ \ v=\int cos(5x+6)dx=\frac{sin(5x+6}{5}[/tex] so now we have:

    [tex]\frac{cos(3x+1)sin(5x+6)}{5}+\frac{3}{5}\int sin(3x+1)sin(5x+6)dx[/tex]

    now for the integral [tex] \int sin(3x+1)sin(5x+6)dx[/tex] take this sub.

    [tex]u=sin(3x+1) \ => du=3cos(3x+1) \ \, and \ \ v=\int sin(5x+6)dx = -\frac{cos(5x+6)}{5}[/tex] now u have:
    [tex] -\frac{sin(3x+1)cos(5x+6)}{5} +\frac{3}{5} \int cos(3x+1)cos(5x+6)dx[/tex]
    Now do you see anything interesting in here, anything that should grab your attention, try to go from here, cuz i think i have almost done it, you are almost there!@!!!!
  6. Feb 28, 2008 #5
    Thanks a lot guys I believe I have come to the right answers. Thanks a million!
  7. Mar 4, 2008 #6
    For number 2.

    How about using the product to sum rule cos(A)cos(B)=1/2[cos(A-B)+cos(A+B)]

    then cos(3x+1)cos(5x+6)=1/2[cos(8x+7)+cos(2x+5)]

    then integrate from here.
  8. Mar 5, 2008 #7
    Not sure if your calculus teacher ever taught you this neat mnemonic, so I'll share it anyways: LIPET, which stands for Logarithmic, Inverse trig., Polynomial, Exponential, and Trigonometric. This is the order in which you should choose for something to set as "u" when integrating by parts. If you've already heard this, sorry I can't be of more help, just thought I'd add on something since you were talking about integration by parts.
  9. Mar 6, 2008 #8
    LIPET is very useful and worth memorizing.
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